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Chemistry

Chemistry Help: Use Mole Ratios

Review real example questions for Use Mole Ratios in Chemistry.

Question 1

Hydrogen peroxide decomposes according to 2H2O2→2H2O+O2\mathrm{2H_2O_2 \rightarrow 2H_2O + O_2}2H2​O2​→2H2​O+O2​. If 7.0 mol7.0\ \text{mol}7.0 mol of H2O2\mathrm{H_2O_2}H2​O2​ decompose completely, how many moles of O2\mathrm{O_2}O2​ form?

  1. 14.0 mol14.0\ \text{mol}14.0 mol
  2. 3.5 mol3.5\ \text{mol}3.5 mol
  3. 7.0 mol7.0\ \text{mol}7.0 mol
  4. 2.0 mol2.0\ \text{mol}2.0 mol
Explanation: This question tests your ability to use mole ratios from balanced equations as conversion factors to calculate how many moles of one substance react with or form from a given number of moles of another substance. Using mole ratios for stoichiometry calculations follows a simple pattern: from the balanced equation, create a conversion factor (fraction) using coefficients where the numerator is the coefficient of the substance you want to find and the denominator is the coefficient of the substance you're given, then multiply the given number of moles by this conversion factor. For example, from 2H2+O2→2H2O2\mathrm{H_2} + \mathrm{O_2} \rightarrow 2\mathrm{H_2O}2H2​+O2​→2H2​O, if given 5 moles of O2\mathrm{O_2}O2​ and asked to find moles of H2O\mathrm{H_2O}H2​O, the mole ratio conversion factor is (2 moles H2O/1 mole O2)(2 \text{ moles } \mathrm{H_2O} / 1 \text{ mole } \mathrm{O_2})(2 moles H2​O/1 mole O2​) from the coefficients, so 5 moles O2×(2 moles H2O/1 mole O2)=10 moles H2O5 \text{ moles } \mathrm{O_2} \times (2 \text{ moles } \mathrm{H_2O} / 1 \text{ mole } \mathrm{O_2}) = 10 \text{ moles } \mathrm{H_2O}5 moles O2​×(2 moles H2​O/1 mole O2​)=10 moles H2​O. The "moles O2\mathrm{O_2}O2​" units cancel, leaving "moles H2O\mathrm{H_2O}H2​O"—dimensional analysis ensures you set up the fraction correctly! For this problem, with 2H2O2→2H2O+O22\mathrm{H_2O_2} \rightarrow 2\mathrm{H_2O} + \mathrm{O_2}2H2​O2​→2H2​O+O2​ and 7.0 mol H2O2\mathrm{H_2O_2}H2​O2​ given to find mol O2\mathrm{O_2}O2​, the conversion factor is (1 mol O2/2 mol H2O2)(1 \text{ mol } \mathrm{O_2} / 2 \text{ mol } \mathrm{H_2O_2})(1 mol O2​/2 mol H2​O2​), so 7.0 mol H2O2×(1/2)=3.5 mol O27.0 \text{ mol } \mathrm{H_2O_2} \times (1/2) = 3.5 \text{ mol } \mathrm{O_2}7.0 mol H2​O2​×(1/2)=3.5 mol O2​. Choice B correctly calculates moles by applying the appropriate coefficient ratio as a conversion factor and performing accurate arithmetic. Something like choice A might happen if you forget to halve and just use 2:1 incorrectly, but always divide by the given coefficient—you're doing great, just double-check the fraction setup! The mole ratio calculation recipe: (1) Write the balanced equation and identify the given substance and wanted substance. (2) Read their coefficients from the equation. (3) Set up conversion factor as fraction: (coefficient of wanted substance / coefficient of given substance). Put what you want on top, what you have on bottom! (4) Multiply: (given moles) × (conversion factor) = answer in moles. Example step-by-step: N2+3H2→2NH3\mathrm{N_2} + 3\mathrm{H_2} \rightarrow 2\mathrm{NH_3}N2​+3H2​→2NH3​. Given: 9 moles H2\mathrm{H_2}H2​. Find: moles NH3\mathrm{NH_3}NH3​. Coefficients: H2\mathrm{H_2}H2​ has 3, NH3\mathrm{NH_3}NH3​ has 2. Conversion factor: (2 moles NH3/3 moles H2)(2 \text{ moles } \mathrm{NH_3} / 3 \text{ moles } \mathrm{H_2})(2 moles NH3​/3 moles H2​). Calculation: 9 moles H2×(2/3)=6 moles NH39 \text{ moles } \mathrm{H_2} \times (2/3) = 6 \text{ moles } \mathrm{NH_3}9 moles H2​×(2/3)=6 moles NH3​. Check: 9:6 simplifies to 3:2, matching coefficient ratio 3:2 for H2\mathrm{H_2}H2​:NH3\mathrm{NH_3}NH3​ ✓. Quick verification trick: after calculating, check if your answer maintains the coefficient ratio. If equation shows 2:1 ratio and you got 6 moles from 3 moles given, does 6:3 equal 2:1? Yes (both are 2:1), so answer likely correct! If equation shows 1:3 ratio and you got 9 moles from 3 moles, does 9:3 equal 1:3? No (9:3 = 3:1, not 1:3), so error occurred—probably inverted the ratio! This proportion check catches most mistakes and takes 3 seconds.

Question 2

Consider the balanced equation: 6CO2+6H2O→C6H12O6+6O26CO_2 + 6H_2O \rightarrow C_6H_{12}O_6 + 6O_26CO2​+6H2​O→C6​H12​O6​+6O2​ If a plant uses 12 mol12 \text{ mol}12 mol of CO2CO_2CO2​ in photosynthesis, how many moles of O2O_2O2​ are produced (assuming enough H2OH_2OH2​O is available)?

  1. 2 mol2 \text{ mol}2 mol
  2. 72 mol72 \text{ mol}72 mol
  3. 6 mol6 \text{ mol}6 mol
  4. 12 mol12 \text{ mol}12 mol
Explanation: This question tests your ability to use mole ratios from balanced equations as conversion factors to calculate how many moles of one substance react with or form from a given number of moles of another substance. Using mole ratios for stoichiometry calculations follows a simple pattern: from the balanced equation, create a conversion factor (fraction) using coefficients where the numerator is the coefficient of the substance you want to find and the denominator is the coefficient of the substance you're given, then multiply the given number of moles by this conversion factor. For example, from 2H2+O2→2H2O2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O}2H2​+O2​→2H2​O, if given 5 moles of O2\text{O}_2O2​ and asked to find moles of H2O\text{H}_2\text{O}H2​O, the mole ratio conversion factor is (2 moles H2O/1 mole O2)(2 \text{ moles } \text{H}_2\text{O} / 1 \text{ mole } \text{O}_2)(2 moles H2​O/1 mole O2​) from the coefficients, so 5 moles O2×(2 moles H2O/1 mole O2)=10 moles H2O5 \text{ moles } \text{O}_2 \times (2 \text{ moles } \text{H}_2\text{O} / 1 \text{ mole } \text{O}_2) = 10 \text{ moles } \text{H}_2\text{O}5 moles O2​×(2 moles H2​O/1 mole O2​)=10 moles H2​O. The "moles O2\text{O}_2O2​" units cancel, leaving "moles H2O\text{H}_2\text{O}H2​O"—dimensional analysis ensures you set up the fraction correctly! For this specific question, with the balanced equation 6CO2+6H2O→C6H12O6+6O26\text{CO}_2 + 6\text{H}_2\text{O} \rightarrow \text{C}_6\text{H}_{12}\text{O}_6 + 6\text{O}_26CO2​+6H2​O→C6​H12​O6​+6O2​ and 12 mol of CO2\text{CO}_2CO2​ given, the conversion factor to find moles of O2\text{O}_2O2​ is (6 mol O2/6 mol CO2)(6 \text{ mol } \text{O}_2 / 6 \text{ mol } \text{CO}_2)(6 mol O2​/6 mol CO2​), so 12 mol CO2×(6/6)=12 mol O212 \text{ mol } \text{CO}_2 \times (6/6) = 12 \text{ mol } \text{O}_212 mol CO2​×(6/6)=12 mol O2​. Choice D correctly calculates the moles by applying the appropriate coefficient ratio as a conversion factor and performing accurate arithmetic. A common distractor like choice B (72 mol) might result from multiplying coefficients unnecessarily, like 6×12, but stick to the ratio and given moles only. The mole ratio calculation recipe: (1) Write the balanced equation and identify the given substance and wanted substance. (2) Read their coefficients from the equation. (3) Set up conversion factor as fraction: (coefficient of wanted substance / coefficient of given substance). Put what you want on top, what you have on bottom! (4) Multiply: (given moles) × (conversion factor) = answer in moles. Example step-by-step: N2+3H2→2NH3\text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3N2​+3H2​→2NH3​. Given: 9 moles H2\text{H}_2H2​. Find: moles NH3\text{NH}_3NH3​. Coefficients: H2\text{H}_2H2​ has 3, NH3\text{NH}_3NH3​ has 2. Conversion factor: (2 moles NH3/3 moles H2)(2 \text{ moles } \text{NH}_3 / 3 \text{ moles } \text{H}_2)(2 moles NH3​/3 moles H2​). Calculation: 9 moles H2×(2/3)=6 moles NH39 \text{ moles } \text{H}_2 \times (2/3) = 6 \text{ moles } \text{NH}_39 moles H2​×(2/3)=6 moles NH3​. Check: 9:6 simplifies to 3:2, matching coefficient ratio 3:2 for H2\text{H}_2H2​: NH3\text{NH}_3NH3​ ✓. Quick verification trick: after calculating, check if your answer maintains the coefficient ratio. If equation shows 2:1 ratio and you got 6 moles from 3 moles given, does 6:3 equal 2:1? Yes (both are 2:1), so answer likely correct! If equation shows 1:3 ratio and you got 9 moles from 3 moles, does 9:3 equal 1:3? No (9:3 = 3:1, not 1:3), so error occurred—probably inverted the ratio! This proportion check catches most mistakes and takes 3 seconds.

Question 3

Aluminum reacts with chlorine gas to form aluminum chloride: 2Al+3Cl2→2AlCl32\text{Al} + 3\text{Cl}_2 \rightarrow 2\text{AlCl}_32Al+3Cl2​→2AlCl3​ If 4.0 mol4.0\ \text{mol}4.0 mol of Al\text{Al}Al react completely, how many moles of Cl2\text{Cl}_2Cl2​ are needed?

  1. 6.0 mol6.0\ \text{mol}6.0 mol
  2. 8.0 mol8.0\ \text{mol}8.0 mol
  3. 3.0 mol3.0\ \text{mol}3.0 mol
  4. 2.0 mol2.0\ \text{mol}2.0 mol
Explanation: This question tests your ability to use mole ratios from balanced equations as conversion factors to calculate how many moles of one substance react with or form from a given number of moles of another substance. Using mole ratios for stoichiometry calculations follows a simple pattern: from the balanced equation, create a conversion factor (fraction) using coefficients where the numerator is the coefficient of the substance you want to find and the denominator is the coefficient of the substance you're given, then multiply the given number of moles by this conversion factor. For example, from 2H2+O2→2H2O2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O}2H2​+O2​→2H2​O, if given 5 moles of O2\text{O}_2O2​ and asked to find moles of H2O\text{H}_2\text{O}H2​O, the mole ratio conversion factor is (2 moles H2O/1 mole O2)(2 \text{ moles } \text{H}_2\text{O} / 1 \text{ mole } \text{O}_2)(2 moles H2​O/1 mole O2​) from the coefficients, so 5 moles O2\text{O}_2O2​ × (2 moles H2O/1 mole O2)(2 \text{ moles } \text{H}_2\text{O} / 1 \text{ mole } \text{O}_2)(2 moles H2​O/1 mole O2​) = 10 moles H2O\text{H}_2\text{O}H2​O. The "moles O2\text{O}_2O2​" units cancel, leaving "moles H2O\text{H}_2\text{O}H2​O"—dimensional analysis ensures you set up the fraction correctly! In this case, for 2Al+3Cl2→2AlCl32\text{Al} + 3\text{Cl}_2 \rightarrow 2\text{AlCl}_32Al+3Cl2​→2AlCl3​ with 4.0 mol Al given and Cl2 wanted, the conversion factor is (3 mol Cl2/2 mol Al)(3 \text{ mol } \text{Cl}_2 / 2 \text{ mol } \text{Al})(3 mol Cl2​/2 mol Al), so 4.0 mol Al × (3/2)(3/2)(3/2) = 6.0 mol Cl2. Choice A correctly calculates moles by applying the appropriate coefficient ratio as a conversion factor and performing accurate arithmetic. Choice B might result from using (2/3)(2/3)(2/3) instead of (3/2)(3/2)(3/2), inverting the ratio, but check by seeing if the proportion matches the coefficients. The mole ratio calculation recipe: (1) Write the balanced equation and identify the given substance and wanted substance. (2) Read their coefficients from the equation. (3) Set up conversion factor as fraction: (coefficient of wanted substance / coefficient of given substance). Put what you want on top, what you have on bottom! (4) Multiply: (given moles) × (conversion factor) = answer in moles. Example step-by-step: N2+3H2→2NH3\text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3N2​+3H2​→2NH3​. Given: 9 moles H2. Find: moles NH3. Coefficients: H2 has 3, NH3 has 2. Conversion factor: (2 moles NH3/3 moles H2)(2 \text{ moles } \text{NH}_3 / 3 \text{ moles } \text{H}_2)(2 moles NH3​/3 moles H2​). Calculation: 9 moles H2 × (2/3)(2/3)(2/3) = 6 moles NH3. Check: 9:6 simplifies to 3:2, matching coefficient ratio 3:2 for H2:NH3 ✓. Quick verification trick: after calculating, check if your answer maintains the coefficient ratio. If equation shows 2:1 ratio and you got 6 moles from 3 moles given, does 6:3 equal 2:1? Yes (both are 2:1), so answer likely correct! If equation shows 1:3 ratio and you got 9 moles from 3 moles, does 9:3 equal 1:3? No (9:3 = 3:1, not 1:3), so error occurred—probably inverted the ratio! This proportion check catches most mistakes and takes 3 seconds.

Question 4

In the balanced reaction: 2KClO3→2KCl+3O22\text{KClO}_3 \rightarrow 2\text{KCl} + 3\text{O}_22KClO3​→2KCl+3O2​ If 4.0 mol4.0\ \text{mol}4.0 mol of KClO3\text{KClO}_3KClO3​ decompose completely, how many moles of O2\text{O}_2O2​ are produced?

  1. 6.0 mol6.0\ \text{mol}6.0 mol
  2. 4.0 mol4.0\ \text{mol}4.0 mol
  3. 2.0 mol2.0\ \text{mol}2.0 mol
  4. 12.0 mol12.0\ \text{mol}12.0 mol
Explanation: This question tests your ability to use mole ratios from balanced equations as conversion factors to calculate how many moles of one substance react with or form from a given number of moles of another substance. Using mole ratios for stoichiometry calculations follows a simple pattern: from the balanced equation, create a conversion factor (fraction) using coefficients where the numerator is the coefficient of the substance you want to find and the denominator is the coefficient of the substance you're given, then multiply the given number of moles by this conversion factor. For example, from 2H2+O2→2H2O2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O}2H2​+O2​→2H2​O, if given 5 moles of O2 and asked to find moles of H2O, the mole ratio conversion factor is (2 moles H2O/1 mole O2)(2 \text{ moles H}_2\text{O} / 1 \text{ mole O}_2)(2 moles H2​O/1 mole O2​) from the coefficients, so 5 moles O2 × (2 moles H2O/1 mole O2)(2 \text{ moles H}_2\text{O} / 1 \text{ mole O}_2)(2 moles H2​O/1 mole O2​) = 10 moles H2O. The "moles O2" units cancel, leaving "moles H2O"—dimensional analysis ensures you set up the fraction correctly! In this case, for 2KClO3→2KCl+3O22\text{KClO}_3 \rightarrow 2\text{KCl} + 3\text{O}_22KClO3​→2KCl+3O2​ with 4.0 mol KClO3 given and O2 wanted, the conversion factor is (3 mol O2/2 mol KClO3)(3 \text{ mol O}_2 / 2 \text{ mol KClO}_3)(3 mol O2​/2 mol KClO3​), so 4.0 mol KClO3 × (3/2)(3/2)(3/2) = 6.0 mol O2. Choice A correctly calculates moles by applying the appropriate coefficient ratio as a conversion factor and performing accurate arithmetic. Choice D might result from mistakenly using (3/1)(3/1)(3/1) or ignoring the 2 for KClO3, but ensure both coefficients are used correctly. The mole ratio calculation recipe: (1) Write the balanced equation and identify the given substance and wanted substance. (2) Read their coefficients from the equation. (3) Set up conversion factor as fraction: (coefficient of wanted substance / coefficient of given substance). Put what you want on top, what you have on bottom! (4) Multiply: (given moles) × (conversion factor) = answer in moles. Example step-by-step: N2+3H2→2NH3N_2 + 3H_2 \rightarrow 2NH_3N2​+3H2​→2NH3​. Given: 9 moles H2. Find: moles NH3. Coefficients: H2 has 3, NH3 has 2. Conversion factor: (2 moles NH3/3 moles H2)(2 \text{ moles NH}_3 / 3 \text{ moles H}_2)(2 moles NH3​/3 moles H2​). Calculation: 9 moles H2 × (2/3)(2/3)(2/3) = 6 moles NH3. Check: 9:6 simplifies to 3:2, matching coefficient ratio 3:2 for H2:NH3 ✓. Quick verification trick: after calculating, check if your answer maintains the coefficient ratio. If equation shows 2:1 ratio and you got 6 moles from 3 moles given, does 6:3 equal 2:1? Yes (both are 2:1), so answer likely correct! If equation shows 1:3 ratio and you got 9 moles from 3 moles, does 9:3 equal 1:3? No (9:3 = 3:1, not 1:3), so error occurred—probably inverted the ratio! This proportion check catches most mistakes and takes 3 seconds.

Question 5

Consider the balanced equation: CaCO3→CaO+CO2CaCO_3 \rightarrow CaO + CO_2CaCO3​→CaO+CO2​ If 7.5 mol7.5 \text{ mol}7.5 mol of CaCO3CaCO_3CaCO3​ decompose completely, how many moles of CO2CO_2CO2​ are produced?

  1. 15.0 mol15.0 \text{ mol}15.0 mol
  2. 7.5 mol7.5 \text{ mol}7.5 mol
  3. 3.75 mol3.75 \text{ mol}3.75 mol
  4. 8.5 mol8.5 \text{ mol}8.5 mol
Explanation: This question tests your ability to use mole ratios from balanced equations as conversion factors to calculate how many moles of one substance react with or form from a given number of moles of another substance. Using mole ratios for stoichiometry calculations follows a simple pattern: from the balanced equation, create a conversion factor (fraction) using coefficients where the numerator is the coefficient of the substance you want to find and the denominator is the coefficient of the substance you're given, then multiply the given number of moles by this conversion factor. For example, from 2H2+O2→2H2O2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O}2H2​+O2​→2H2​O, if given 5 moles of O2\text{O}_2O2​ and asked to find moles of H2O\text{H}_2\text{O}H2​O, the mole ratio conversion factor is (2 moles H2O/1 mole O2)(2 \text{ moles } \text{H}_2\text{O} / 1 \text{ mole } \text{O}_2)(2 moles H2​O/1 mole O2​) from the coefficients, so 5 moles O2\text{O}_2O2​ × (2 moles H2O/1 mole O2)(2 \text{ moles } \text{H}_2\text{O} / 1 \text{ mole } \text{O}_2)(2 moles H2​O/1 mole O2​) = 10 moles H2O\text{H}_2\text{O}H2​O. The "moles O2\text{O}_2O2​" units cancel, leaving "moles H2O\text{H}_2\text{O}H2​O"—dimensional analysis ensures you set up the fraction correctly! For this specific question, with the balanced equation CaCO3→CaO+CO2\text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2CaCO3​→CaO+CO2​ and 7.5 mol of CaCO3\text{CaCO}_3CaCO3​ given, the conversion factor to find moles of CO2\text{CO}_2CO2​ is (1 mol CO2/1 mol CaCO3)(1 \text{ mol } \text{CO}_2 / 1 \text{ mol } \text{CaCO}_3)(1 mol CO2​/1 mol CaCO3​), so 7.5 mol CaCO3\text{CaCO}_3CaCO3​ × (1/1)(1/1)(1/1) = 7.5 mol CO2\text{CO}_2CO2​. Choice B correctly calculates the moles by applying the appropriate coefficient ratio as a conversion factor and performing accurate arithmetic. A common distractor like choice A (15.0 mol) might result from doubling the amount or using a wrong coefficient, but confirm it's a 1:1 ratio here. The mole ratio calculation recipe: (1) Write the balanced equation and identify the given substance and wanted substance. (2) Read their coefficients from the equation. (3) Set up conversion factor as fraction: (coefficient of wanted substance / coefficient of given substance). Put what you want on top, what you have on bottom! (4) Multiply: (given moles) × (conversion factor) = answer in moles. Example step-by-step: N2+3H2→2NH3\text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3N2​+3H2​→2NH3​. Given: 9 moles H2\text{H}_2H2​. Find: moles NH3\text{NH}_3NH3​. Coefficients: H2\text{H}_2H2​ has 3, NH3\text{NH}_3NH3​ has 2. Conversion factor: (2 moles NH3/3 moles H2)(2 \text{ moles } \text{NH}_3 / 3 \text{ moles } \text{H}_2)(2 moles NH3​/3 moles H2​). Calculation: 9 moles H2\text{H}_2H2​ × (2/3)(2/3)(2/3) = 6 moles NH3\text{NH}_3NH3​. Check: 9:6 simplifies to 3:2, matching coefficient ratio 3:2 for H2\text{H}_2H2​: NH3\text{NH}_3NH3​ ✓. Quick verification trick: after calculating, check if your answer maintains the coefficient ratio. If equation shows 2:1 ratio and you got 6 moles from 3 moles given, does 6:3 equal 2:1? Yes (both are 2:1), so answer likely correct! If equation shows 1:3 ratio and you got 9 moles from 3 moles, does 9:3 equal 1:3? No (9:3 = 3:1, not 1:3), so error occurred—probably inverted the ratio! This proportion check catches most mistakes and takes 3 seconds.

Question 6

Consider the balanced equation: C3H8+5O2→3CO2+4H2OC_3H_8 + 5O_2 \rightarrow 3CO_2 + 4H_2OC3​H8​+5O2​→3CO2​+4H2​O If 2.0 mol2.0\ \text{mol}2.0 mol of C3H8C_3H_8C3​H8​ burn completely, how many moles of CO2CO_2CO2​ are produced?

  1. 23 mol\tfrac{2}{3}\ \text{mol}32​ mol
  2. 6.0 mol6.0\ \text{mol}6.0 mol
  3. 10.0 mol10.0\ \text{mol}10.0 mol
  4. 3.0 mol3.0\ \text{mol}3.0 mol
Explanation: This question tests your ability to use mole ratios from balanced equations as conversion factors to calculate how many moles of one substance react with or form from a given number of moles of another substance. Using mole ratios for stoichiometry calculations follows a simple pattern: from the balanced equation, create a conversion factor (fraction) using coefficients where the numerator is the coefficient of the substance you want to find and the denominator is the coefficient of the substance you're given, then multiply the given number of moles by this conversion factor. For example, from 2H2+O2→2H2O2H_2 + O_2 \rightarrow 2H_2O2H2​+O2​→2H2​O, if given 5 moles of O2 and asked to find moles of H2O, the mole ratio conversion factor is (2 moles H2O/1 mole O2)(2 \text{ moles H}_2\text{O} / 1 \text{ mole O}_2)(2 moles H2​O/1 mole O2​) from the coefficients, so 5 moles O2 × (2 moles H2O/1 mole O2)(2 \text{ moles H}_2\text{O} / 1 \text{ mole O}_2)(2 moles H2​O/1 mole O2​) = 10 moles H2O. The "moles O2" units cancel, leaving "moles H2O"—dimensional analysis ensures you set up the fraction correctly! For this specific question, with the balanced equation C3H8+5O2→3CO2+4H2OC_3H_8 + 5O_2 \rightarrow 3CO_2 + 4H_2OC3​H8​+5O2​→3CO2​+4H2​O and 2.0 mol of C3H8 given, the conversion factor to find moles of CO2 is (3 mol CO2/1 mol C3H8)(3 \text{ mol CO}_2 / 1 \text{ mol C}_3\text{H}_8)(3 mol CO2​/1 mol C3​H8​), so 2.0 mol C3H8 × (3/1)(3/1)(3/1) = 6.0 mol CO2. Choice B correctly calculates the moles by applying the appropriate coefficient ratio as a conversion factor and performing accurate arithmetic. A common distractor like choice A (23 mol\tfrac{2}{3} \text{ mol}32​ mol) might result from inverting the ratio to (1/3) or a calculation error, but always put the wanted on top to avoid inversion mistakes. The mole ratio calculation recipe: (1) Write the balanced equation and identify the given substance and wanted substance. (2) Read their coefficients from the equation. (3) Set up conversion factor as fraction: (coefficient of wanted substance / coefficient of given substance). Put what you want on top, what you have on bottom! (4) Multiply: (given moles) × (conversion factor) = answer in moles. Example step-by-step: N2+3H2→2NH3N_2 + 3H_2 \rightarrow 2NH_3N2​+3H2​→2NH3​. Given: 9 moles H2. Find: moles NH3. Coefficients: H2 has 3, NH3 has 2. Conversion factor: (2 moles NH3/3 moles H2)(2 \text{ moles NH}_3 / 3 \text{ moles H}_2)(2 moles NH3​/3 moles H2​). Calculation: 9 moles H2 × (2/3)(2/3)(2/3) = 6 moles NH3. Check: 9:6 simplifies to 3:2, matching coefficient ratio 3:2 for H2:NH3 ✓. Quick verification trick: after calculating, check if your answer maintains the coefficient ratio. If equation shows 2:1 ratio and you got 6 moles from 3 moles given, does 6:3 equal 2:1? Yes (both are 2:1), so answer likely correct! If equation shows 1:3 ratio and you got 9 moles from 3 moles, does 9:3 equal 1:3? No (9:3 = 3:1, not 1:3), so error occurred—probably inverted the ratio! This proportion check catches most mistakes and takes 3 seconds.

Question 7

Iron reacts with oxygen to form iron(III) oxide: 4Fe+3O2→2Fe2O34\text{Fe} + 3\text{O}_2 \rightarrow 2\text{Fe}_2\text{O}_34Fe+3O2​→2Fe2​O3​ If 8.0 mol8.0\ \text{mol}8.0 mol of Fe\text{Fe}Fe react completely, how many moles of Fe2O3\text{Fe}_2\text{O}_3Fe2​O3​ are produced?

  1. 16.0mol16.0 \text{mol}16.0mol
  2. 6.0mol6.0 \text{mol}6.0mol
  3. 4.0mol4.0 \text{mol}4.0mol
  4. 8.0mol8.0 \text{mol}8.0mol
Explanation: This question tests your ability to use mole ratios from balanced equations as conversion factors to calculate how many moles of one substance react with or form from a given number of moles of another substance. Using mole ratios for stoichiometry calculations follows a simple pattern: from the balanced equation, create a conversion factor (fraction) using coefficients where the numerator is the coefficient of the substance you want to find and the denominator is the coefficient of the substance you're given, then multiply the given number of moles by this conversion factor. For example, from 2H2+O2→2H2O2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O}2H2​+O2​→2H2​O, if given 5 moles of O2\text{O}_2O2​ and asked to find moles of H2O\text{H}_2\text{O}H2​O, the mole ratio conversion factor is (2 moles H2O/1 mole O2)(2 \text{ moles } \text{H}_2\text{O} / 1 \text{ mole } \text{O}_2)(2 moles H2​O/1 mole O2​) from the coefficients, so 5 moles O2\text{O}_2O2​ × (2 moles H2O/1 mole O2)(2 \text{ moles } \text{H}_2\text{O} / 1 \text{ mole } \text{O}_2)(2 moles H2​O/1 mole O2​) = 10 moles H2O\text{H}_2\text{O}H2​O. The "moles O2\text{O}_2O2​" units cancel, leaving "moles H2O\text{H}_2\text{O}H2​O"—dimensional analysis ensures you set up the fraction correctly! In this case, for 4Fe+3O2→2Fe2O34\text{Fe} + 3\text{O}_2 \rightarrow 2\text{Fe}_2\text{O}_34Fe+3O2​→2Fe2​O3​ with 8.0 mol Fe given and Fe2O3\text{Fe}_2\text{O}_3Fe2​O3​ wanted, the conversion factor is (2 mol Fe2O3/4 mol Fe)(2 \text{ mol } \text{Fe}_2\text{O}_3 / 4 \text{ mol } \text{Fe})(2 mol Fe2​O3​/4 mol Fe), so 8.0 mol Fe × (2/4)(2/4)(2/4) = 4.0 mol Fe2O3\text{Fe}_2\text{O}_3Fe2​O3​. Choice C correctly calculates moles by applying the appropriate coefficient ratio as a conversion factor and performing accurate arithmetic. Choice A might result from multiplying by (4/2)(4/2)(4/2) instead of (2/4)(2/4)(2/4), but always verify the ratio direction with the equation's coefficients. The mole ratio calculation recipe: (1) Write the balanced equation and identify the given substance and wanted substance. (2) Read their coefficients from the equation. (3) Set up conversion factor as fraction: (coefficient of wanted substance / coefficient of given substance). Put what you want on top, what you have on bottom! (4) Multiply: (given moles) × (conversion factor) = answer in moles. Example step-by-step: N2+3H2→2NH3\text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3N2​+3H2​→2NH3​. Given: 9 moles H2\text{H}_2H2​. Find: moles NH3\text{NH}_3NH3​. Coefficients: H2\text{H}_2H2​ has 3, NH3\text{NH}_3NH3​ has 2. Conversion factor: (2 moles NH3/3 moles H2)(2 \text{ moles } \text{NH}_3 / 3 \text{ moles } \text{H}_2)(2 moles NH3​/3 moles H2​). Calculation: 9 moles H2\text{H}_2H2​ × (2/3)(2/3)(2/3) = 6 moles NH3\text{NH}_3NH3​. Check: 9:6 simplifies to 3:2, matching coefficient ratio 3:2 for H2\text{H}_2H2​: NH3\text{NH}_3NH3​ ✓. Quick verification trick: after calculating, check if your answer maintains the coefficient ratio. If equation shows 2:1 ratio and you got 6 moles from 3 moles given, does 6:3 equal 2:1? Yes (both are 2:1), so answer likely correct! If equation shows 1:3 ratio and you got 9 moles from 3 moles, does 9:3 equal 1:3? No (9:3 = 3:1, not 1:3), so error occurred—probably inverted the ratio! This proportion check catches most mistakes and takes 3 seconds.

Question 8

Iron reacts with oxygen to form iron(III) oxide: 4Fe+3O2→2Fe2O3\textbf{4Fe} + \textbf{3O}_2 \rightarrow \textbf{2Fe}_2\textbf{O}_34Fe+3O2​→2Fe2​O3​ If 6.0 mol6.0\ \text{mol}6.0 mol of Fe\text{Fe}Fe react completely, how many moles of Fe2O3\text{Fe}_2\text{O}_3Fe2​O3​ form?

  1. 3.0 mol3.0\ \text{mol}3.0 mol
  2. 12.0 mol12.0\ \text{mol}12.0 mol
  3. 4.5 mol4.5\ \text{mol}4.5 mol
  4. 1.5 mol1.5\ \text{mol}1.5 mol
Explanation: This question tests your ability to use mole ratios from balanced equations as conversion factors to calculate how many moles of one substance react with or form from a given number of moles of another substance. Using mole ratios for stoichiometry calculations follows a simple pattern: from the balanced equation, create a conversion factor (fraction) using coefficients where the numerator is the coefficient of the substance you want to find and the denominator is the coefficient of the substance you're given, then multiply the given number of moles by this conversion factor. For the equation 4Fe + 3O₂ → 2Fe₂O₃, we're given 6.0 mol Fe and need to find moles of Fe₂O₃ formed. The mole ratio conversion factor is (2 moles Fe₂O₃ / 4 moles Fe) from the coefficients, so: 6.0 mol Fe × (2 mol Fe₂O₃ / 4 mol Fe) = 6.0 × (2/4) = 6.0 × (1/2) = 3.0 mol Fe₂O₃. Choice A correctly calculates 3.0 mol by applying the 2:4 (or 1:2) coefficient ratio as a conversion factor and performing accurate arithmetic. Common errors include using the wrong ratio or multiplying by 2 instead of dividing, which would give 12.0 mol. The mole ratio calculation recipe: (1) Write the balanced equation and identify the given substance and wanted substance. (2) Read their coefficients from the equation. (3) Set up conversion factor as fraction: (coefficient of wanted substance / coefficient of given substance). Put what you want on top, what you have on bottom! (4) Multiply: (given moles) × (conversion factor) = answer in moles. Quick verification: 6.0 mol Fe produces 3.0 mol Fe₂O₃, giving ratio 6:3 = 2:1, matching the coefficient ratio 4:2 = 2:1 for Fe:Fe₂O₃ ✓.

Question 9

Iron reacts with oxygen to form iron(III) oxide: 4Fe+3O2→2Fe2O3\mathrm{4Fe + 3O_2 \rightarrow 2Fe_2O_3}4Fe+3O2​→2Fe2​O3​ If 8.0 mol8.0\ \text{mol}8.0 mol of Fe\mathrm{Fe}Fe react completely, how many moles of Fe2O3\mathrm{Fe_2O_3}Fe2​O3​ form?

  1. 16.0 mol16.0\ \text{mol}16.0 mol
  2. 6.0 mol6.0\ \text{mol}6.0 mol
  3. 4.0 mol4.0\ \text{mol}4.0 mol
  4. 8.0 mol8.0\ \text{mol}8.0 mol
Explanation: This question tests your ability to use mole ratios from balanced equations as conversion factors to calculate how many moles of one substance react with or form from a given number of moles of another substance. Using mole ratios for stoichiometry calculations follows a simple pattern: from the balanced equation, create a conversion factor (fraction) using coefficients where the numerator is the coefficient of the substance you want to find and the denominator is the coefficient of the substance you're given, then multiply the given number of moles by this conversion factor. For example, from 2H2+O2→2H2O2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O}2H2​+O2​→2H2​O, if given 5 moles of O2 and asked to find moles of H2O, the mole ratio conversion factor is (2 moles H2O/1 mole O22 \text{ moles H}_2\text{O} / 1 \text{ mole O}_22 moles H2​O/1 mole O2​) from the coefficients, so 5 moles O2×(2 moles H2O/1 mole O2)=10 moles H2O5 \text{ moles O}_2 \times (2 \text{ moles H}_2\text{O} / 1 \text{ mole O}_2) = 10 \text{ moles H}_2\text{O}5 moles O2​×(2 moles H2​O/1 mole O2​)=10 moles H2​O. The "moles O2" units cancel, leaving "moles H2O"—dimensional analysis ensures you set up the fraction correctly! For this problem, with 4Fe+3O2→2Fe2O34\text{Fe} + 3\text{O}_2 \rightarrow 2\text{Fe}_2\text{O}_34Fe+3O2​→2Fe2​O3​ and 8.0 mol Fe given to find mol Fe2O3, the conversion factor is (2 mol Fe2O3/4 mol Fe2 \text{ mol Fe}_2\text{O}_3 / 4 \text{ mol Fe}2 mol Fe2​O3​/4 mol Fe), so 8.0 mol Fe×(2/4)=4.0 mol Fe2O38.0 \text{ mol Fe} \times (2/4) = 4.0 \text{ mol Fe}_2\text{O}_38.0 mol Fe×(2/4)=4.0 mol Fe2​O3​. Choice C correctly calculates moles by applying the appropriate coefficient ratio as a conversion factor and performing accurate arithmetic. A distractor like choice D could stem from using only half the ratio or forgetting to simplify 2/4 to 1/2, but practice makes perfect—ensure the coefficients match exactly! The mole ratio calculation recipe: (1) Write the balanced equation and identify the given substance and wanted substance. (2) Read their coefficients from the equation. (3) Set up conversion factor as fraction: (coefficient of wanted substance / coefficient of given substance). Put what you want on top, what you have on bottom! (4) Multiply: (given moles) × (conversion factor) = answer in moles. Example step-by-step: N2+3H2→2NH3\text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3N2​+3H2​→2NH3​. Given: 9 moles H2. Find: moles NH3. Coefficients: H2 has 3, NH3 has 2. Conversion factor: (2 moles NH3/3 moles H22 \text{ moles NH}_3 / 3 \text{ moles H}_22 moles NH3​/3 moles H2​). Calculation: 9 moles H2×(2/3)=6 moles NH39 \text{ moles H}_2 \times (2/3) = 6 \text{ moles NH}_39 moles H2​×(2/3)=6 moles NH3​. Check: 9:6 simplifies to 3:2, matching coefficient ratio 3:2 for H2:NH3 ✓. Quick verification trick: after calculating, check if your answer maintains the coefficient ratio. If equation shows 2:1 ratio and you got 6 moles from 3 moles given, does 6:3 equal 2:1? Yes (both are 2:1), so answer likely correct! If equation shows 1:3 ratio and you got 9 moles from 3 moles, does 9:3 equal 1:3? No (9:3 = 3:1, not 1:3), so error occurred—probably inverted the ratio! This proportion check catches most mistakes and takes 3 seconds.

Question 10

Aluminum reacts with chlorine to form aluminum chloride: 2Al+3Cl2→2AlCl3\mathrm{2Al + 3Cl_2 \rightarrow 2AlCl_3}2Al+3Cl2​→2AlCl3​ If 3.0 mol3.0 \, \text{mol}3.0mol of Cl2\mathrm{Cl_2}Cl2​ react completely, how many moles of AlCl3\mathrm{AlCl_3}AlCl3​ form?

  1. 2.0 mol2.0 \, \text{mol}2.0mol
  2. 4.5 mol4.5 \, \text{mol}4.5mol
  3. 1.5 mol1.5 \, \text{mol}1.5mol
  4. 3.0 mol3.0 \, \text{mol}3.0mol
Explanation: This question tests your ability to use mole ratios from balanced equations as conversion factors to calculate how many moles of one substance react with or form from a given number of moles of another substance. Using mole ratios for stoichiometry calculations follows a simple pattern: from the balanced equation, create a conversion factor (fraction) using coefficients where the numerator is the coefficient of the substance you want to find and the denominator is the coefficient of the substance you're given, then multiply the given number of moles by this conversion factor. For example, from 2H2+O2→2H2O2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O}2H2​+O2​→2H2​O, if given 5 moles of O2 and asked to find moles of H2O, the mole ratio conversion factor is (2 moles H2O1 mole O2\frac{2 \text{ moles H}_2\text{O}}{1 \text{ mole O}_2}1 mole O2​2 moles H2​O​) from the coefficients, so 5 moles O2 × (2 moles H2O1 mole O2\frac{2 \text{ moles H}_2\text{O}}{1 \text{ mole O}_2}1 mole O2​2 moles H2​O​) = 10 moles H2O. The "moles O2" units cancel, leaving "moles H2O"—dimensional analysis ensures you set up the fraction correctly! For this problem, with 2Al+3Cl2→2AlCl32\text{Al} + 3\text{Cl}_2 \rightarrow 2\text{AlCl}_32Al+3Cl2​→2AlCl3​ and 3.0 mol Cl2 given to find mol AlCl3, the conversion factor is (2 mol AlCl33 mol Cl2\frac{2 \text{ mol AlCl}_3}{3 \text{ mol Cl}_2}3 mol Cl2​2 mol AlCl3​​), so 3.0 mol Cl2 × (23\frac{2}{3}32​) = 2.0 mol AlCl3. Choice A correctly calculates moles by applying the appropriate coefficient ratio as a conversion factor and performing accurate arithmetic. Choice C could result from using 32\frac{3}{2}23​ instead, inverting the fraction, but always verify with the proportion check to catch that—excellent effort! The mole ratio calculation recipe: (1) Write the balanced equation and identify the given substance and wanted substance. (2) Read their coefficients from the equation. (3) Set up conversion factor as fraction: (coefficient of wanted substancecoefficient of given substance\frac{\text{coefficient of wanted substance}}{\text{coefficient of given substance}}coefficient of given substancecoefficient of wanted substance​). Put what you want on top, what you have on bottom! (4) Multiply: (given moles) × (conversion factor) = answer in moles. Example step-by-step: N2+3H2→2NH3\text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3N2​+3H2​→2NH3​. Given: 9 moles H2. Find: moles NH3. Coefficients: H2 has 3, NH3 has 2. Conversion factor: (2 moles NH33 moles H2\frac{2 \text{ moles NH}_3}{3 \text{ moles H}_2}3 moles H2​2 moles NH3​​). Calculation: 9 moles H2 × (23\frac{2}{3}32​) = 6 moles NH3. Check: 9:6 simplifies to 3:2, matching coefficient ratio 3:2 for H2:NH3 ✓. Quick verification trick: after calculating, check if your answer maintains the coefficient ratio. If equation shows 2:1 ratio and you got 6 moles from 3 moles given, does 6:3 equal 2:1? Yes (both are 2:1), so answer likely correct! If equation shows 1:3 ratio and you got 9 moles from 3 moles, does 9:3 equal 1:3? No (9:3 = 3:1, not 1:3), so error occurred—probably inverted the ratio! This proportion check catches most mistakes and takes 3 seconds.