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Example Questions

Example Question #3 : Balancing Chemical Equations

_Fe₂O₃ + _HCl ⇌ _FeCl₃ + _H₂O

The Following question will be based on the unbalanced reaction above

For the reaction above to be balanced what coefficient should be in front of the compound HCl?

Possible Answers:

Correct answer:

Explanation:

Balancing reactions is best acheived by using a stepwise approach. It is useful to work through each atom making sure it is present in a balanced fashion on both sides of the equation. Starting with Fe, Fe₂O₃is the only molecule with Fe present on the left side of the equation and FeCl₃ is the only molecule with Fe present on the right side of the equation. Thus the molar ratio of Fe₂O₃to FeCl₃ must be 1:2. Moving on to O, the O on the left side of the equation exists as Fe₂O₃ and H₂O on the right. The molar ratio of Fe₂O₃to H₂O is therefore 1:3. Cl exists in HCl on the left and FeCl₃on the right. Thus the molar ratio of HCl to FeCl₃ must be 3:1. To ensure that these molar ratios are maintaned the balanced formula is then determined to be Fe₂O₃ + 6HCl -> 2FeCl₃ + 3H₂O

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Example Question #1 : Balancing Equations

Balance the following equation:

FeCl₃+ NOH₅→ Fe(OH)₃ + NH₄Cl

Possible Answers:

FeCl₃ + 3 NOH₅ → Fe(OH)₃ + 3 NH₄Cl

3 FeCl₃ + 3 NOH₅ → 2 Fe(OH)₃ + 4 NH₄Cl

2 FeCl₃ + 3 NOH₅ → 2 Fe(OH)₃ + 3 NH₄Cl

2 FeCl₃ + 4 NOH₅ → Fe(OH)₃ + 6 NH₄Cl

3 FeCl₃ + 3 NOH₅ → 3 Fe(OH)₃ + 3 NH₄Cl

Correct answer:

FeCl₃ + 3 NOH₅ → Fe(OH)₃ + 3 NH₄Cl

Explanation:

The first thing to do is to balance the Cl (3 on the left; one on the right) ← add 3 for NH₄Cl

Now, there are 3 N on the right and only one on the left; add a 3 to the NOH₅

Check to see that the Fe, H, and O balance, which they do

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Example Question #2 : Balancing Equations

After the following reaction is balanced, how many moles of H₂O can 4 moles of C₂H₆produce?

__C₂H₆(s) + ___O₂(g) → ___H₂O(l) + ___CO₂(g)

Possible Answers:

Correct answer:

Explanation:

The reaction balances out to 2C₂H₆(s) + 7O₂(g) → 6H₂O(l) + 4CO₂(g). For every 2 moles of C₂H₆ you can produce 6 moles of H₂O. Giving you a total production of 12 moles when you have 4 moles of C₂H₆.

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Example Question #51 : Chemical Reactions

Consider the following reaction:

$KMnO_{4} + HCl \rightarrow KCl + MnCl_{2} + H_{2}O + Cl_{2}$

When the equation is balanced, what will be the coefficient in front of HCl?

Possible Answers:

Correct answer:

Explanation:

When balancing equations, the goal is to make sure that the same atoms, in both type and amount, are on both the reactant and product side of the equation. A helpful approach is to write down the number of atoms already on both sides of the unbalanced equation. This way, you can predict which compounds need to be increased on which side in order to balance the equation. It also helps to balance oxygen and hydrogen last in the equation.

In this reaction, we can balance as follows.

Reactants: 1K, 1Mn, 1Cl, 4O, 1H

Products: 1K, 1Mn, 5Cl, 1O, 2H

So, we will need to increase H₂O and HCl. The final balanced equation is written below.

$2KMnO_{4} + 16HCl \rightarrow 2KCl + 2MnCl_{2} + 8H_{2}O + 5Cl_{2}$

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Example Question #52 : Chemical Reactions

Balance the following chemical equation.

$NaOH + FeCl_{3} \rightarrow NaCl + Fe(OH)_3$

Possible Answers:

$3NaOH+2FeCl_3\rightarrow 2NaCl+2Fe(OH)_3$

$NaOH + FeCl_{3} \rightarrow 3NaCl + Fe(OH)_3$

$3NaOH + FeCl_{3} \rightarrow 3NaCl + Fe(OH)_3$

$3NaOH + FeCl_{3} \rightarrow NaCl + Fe(OH)_3$

$3NaOH+FeCl_3\rightarrow 3NaCl+3Fe(OH)_3$

Correct answer:

$3NaOH + FeCl_{3} \rightarrow 3NaCl + Fe(OH)_3$

Explanation:

To balance an equation, we need to make sure there is the same amount of elements to the left of the arrow as there is to the right. We also need all the charges to balance out. We notice right away that there are three chlorine atoms on the left, but only one on the right.

$NaOH + FeCl_{3} \rightarrow NaCl + Fe(OH)_3$ (1Na, 1O, 1H, 1Fe, 3Cl : 1Na, 3O, 3H, 1Fe, 1Cl)

We can solve this by multiplying NaCl by three.

$NaOH + FeCl_{3} \rightarrow 3NaCl + Fe(OH)_3$ (1Na, 1O, 1H, 1Fe, 3Cl : 3Na, 3O, 3H, 1Fe, 3Cl)

This causes us to have an imbalance of sodium, which we can correct by manipulating NaOH.

$3NaOH + FeCl_{3} \rightarrow 3NaCl + Fe(OH)_3$ (3Na, 3O, 3H, 1Fe, 3Cl : 3Na, 3O, 3H, 1Fe, 1Cl)

This is the final balanced equation. Note that it is usually easiest to manipulate oxygen and hydrogen last, since they are often involved in multiple molecules.

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Example Question #1 : Balancing Chemical Equations

Calcium hydroxide is treated with hydrochloric acid to produce water and calcium chloride. Write a balanced chemical reaction that describes this process.

Possible Answers:

$Ca(OH)_{2} + HCl \rightarrow H_{2}O + CaCl_{2}$

$Ca(OH)_{3} + 3HCl \rightarrow 3H_{2}O + CaCl_{3}$

$Ca(OH)_{2} + 2HCl \rightarrow 2H_{2}O + CaCl_{2}$

$CaOH + HCl \rightarrow H_{2}O + CaCl$

$2Ca(OH)_{2} + HCl \rightarrow H_{2}O + 2CaCl_{2}$

Correct answer:

$Ca(OH)_{2} + 2HCl \rightarrow 2H_{2}O + CaCl_{2}$

Explanation:

Calcium is in the second group of the periodic table, and is therefore going to have a $\small +2$ oxidation number. Hydroxide ions have a $\small -1$ charge. Calcium hydroxide will have the formula $\small Ca(OH)_2$ .

Chloride ions have a $\small -1$ charge and hydrogen ions have a $\small +1$ charge. The formula for hydrochloric acid is $\small HCl$ .

On the products side, water has the formula $\small H_2O$ and calcium chloride has the formula $\small CaCl_2$ .

Now that we know all of the formulas, we can write our reaction:

$Ca(OH)_2+HCl\rightarrow H_2O+CaCl_2$

In order to balance the chloride atoms, we need to add coefficients.

$Ca(OH)_2+2HCl\rightarrow 2H_2O+CaCl_2$

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Example Question #53 : Chemical Reactions

Consider the following unbalanced equation for the combustion of propane, $C_3H_8$ :

$C_3H_{8\hspace{1 mm}(g)}+O_{2\hspace{1 mm}(g)}\rightarrow CO_{2\hspace{1 mm}(g)}+H_2O_{(l)}$

If you were to combust one mole of propane, how many moles of water would you produce?

Possible Answers:

$3 mol\ H_2O$

$1 mol\ H_2O$

$5 mol\ H_2O$

$4 mol\ H_2O$

$2 mol\ H_2O$

Correct answer:

$4 mol\ H_2O$

Explanation:

Begin by balancing the equation. There are many ways to do this, but one method that is particularly useful is to assume that you have 1 mole of your hydrocarbon (propane), and balance the equation from there. It may be necessary to manipulate an equation further if you end up with fractions, but all you will need to do is multiply by an integer if that is the case.

First, we will balance the carbons. There are three carbons in propane, so we will make sure there are three carbons on the right side of the arrow as well:

$C_3H_8 + O_2 \rightarrow 3CO_2 + H_2O$

This is not complete. We will next balance the hydrogens. There are eight hydrogens on the left side of the equation, so:

$C_3H_8 + O_2 \rightarrow 3CO_2 + 4H_2O$

The last step is to balance the oxygens on the left and right side of the equation

$C_3H_8 + 5O_2 \rightarrow 3CO_2 + 4H_2O$

Our equation is balanced and all coefficients are integers. If we begin with one mole of propane, we will produce four moles of water.

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Example Question #54 : Chemical Reactions

Consider the following unbalanced equation:

$Fe+Cl_2\rightarrow FeCl_3$

How many grams of solid iron are needed to make 36.0g of $FeCl_3$ ? Assume that chlorine is in excess.

Possible Answers:

$18.6\hspace{1 mm}g\hspace{1 mm}Fe$

$8.27\hspace{1 mm}g\hspace{1 mm}Fe$

$36.0\hspace{1 mm}g\hspace{1 mm}Fe$

$2.22*10^{-1}g\ Fe$

$12.4\hspace{1 mm}g\hspace{1 mm}Fe$

Correct answer:

$12.4\hspace{1 mm}g\hspace{1 mm}Fe$

Explanation:

$Fe+Cl_2\rightarrow FeCl_3$

First, we will balance the equation:

$2Fe+3Cl_2\rightarrow 2FeCl_3$

Since chlorine is in excess, we know that the limiting reagent is iron.

$36.0\hspace{1 mm}g\hspace{1 mm}FeCl_3\times\frac{1\hspace{1 mm}mole\hspace{1 mm}FeCl_3}{162.2\hspace{1 mm}g\hspace{1 mm}FeCl_3}\times\frac{2\hspace{1 mm}moles\hspace{1 mm}Fe}{2\hspace{1 mm}moles\hspace{1 mm}FeCl_3}\times\frac{55.845\hspace{1 mm}g\hspace{1 mm}Fe}{1\hspace{1 mm}mole\hspace{1 mm}Fe}=12.4\hspace{1 mm}g\hspace{1 mm}Fe$

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Example Question #1 : Balancing Equations

$S + HNO_{3} \rightleftharpoons H_{2}SO_{4} + NO_{2} + H_{2}O$

In the balanced version of the preceding equation, what is the coefficient of nitrogen dioxide?

Possible Answers:

Correct answer:

Explanation:

In the balanced version of the equation:

$S + 6HNO_{3} \rightleftharpoons H_{2}SO_{4} + 6NO_{2} + 2H_{2}O$

Nitrogen dioxide's coefficient is 6 in order to balance the 6 moles of nitrogen provided by the 6 moles of nitric acid on the equation's left side.

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Example Question #11 : Stoichiometry

Transesterification is an industrially important process for the production of biodiesel fuel (that most diesel engines can run cleanly on) and glycerin from triglycerides (usually people use vegetable oil as their feedstock).

One easy to do and high yielding method involves the reaction of ethanol or methanol with sodium hydroxide to produce the super bases sodium ethoxide or sodium methoxide respectively. The super base is then used in the transesterification reaction with triglycerides to produce glycerin and ethyl or methyl esters (biodiesel). The products can easily be separated by differences in density. Note: For this reaction to occur the reactants must be free of water as when water is present saponification occurs (how soap is made) which will compete with the transesterification reaction. Note: just because the reaction described above is easy doesn't mean that it is anywhere near safe; please do thoroughly research any reaction you plan to undertake and take all possible safety precautions. That being said, this process is becoming much more popular lately and with the recent focus on alternative energy may well soon account for a decent portion of fuel production.

Catalyst formation equation:

$NaOH+CH_3OH\rightleftharpoons Na^+CH_3OH+H_2O$

Unbalanced transesterification equation:

$RCO_2CH_2CH(O_2CR)CH_2CO_2R +CH_3OH \xrightarrow[]{Catalyst} C_3H_8O_3+CH_3CO_2R$

Balance the following transesterification reaction assuming that all R-groups are identical:

$RCO_2CH_2CH(O_2CR)CH_2CO_2R +CH_3OH \xrightarrow[]{Catalyst} C_3H_8O_3+CH_3CO_2R$

Possible Answers:

$3RCO_2CH_2CH(O_2CR)CH_2CO_2R +2CH_3OH \xrightarrow[]{Catalyst} 5C_3H_8O_3+CH_3CO_2R$

$RCO_2CH_2CH(O_2CR)CH_2CO_2R +CH_3OH \xrightarrow[]{Catalyst} C_3H_8O_3+CH_3CO_2R$

$RCO_2CH_2CH(O_2CR)CH_2CO_2R +3CH_3OH \xrightarrow[]{Catalyst} C_3H_8O_3+3CH_3CO_2R$

$2RCO_2CH_2CH(O_2CR)CH_2CO_2R +CH_3OH \xrightarrow[]{Catalyst} C_3H_8O_3+4CH_3CO_2R$

$\frac{1}{2}RCO_2CH_2CH(O_2CR)CH_2CO_2R +CH_3OH \xrightarrow[]{Catalyst} C_3H_8O_3+2CH_3CO_2R$

Correct answer:

$RCO_2CH_2CH(O_2CR)CH_2CO_2R +3CH_3OH \xrightarrow[]{Catalyst} C_3H_8O_3+3CH_3CO_2R$

Explanation:

The steps for balancing equations are as follows and should be done in order:

1.) Check for diatomic molecules: In this case there are none so move on to step 2

2.) Balance the metals (hydrogen is not considered a metal in this application): We have no metals in our equation, so move on to step 3 (the sodium is part of the catalyst so we don't consider it part of the balancing).

3.) Balance the nonmetals (not including oxygen). If you notice the left hand side of the equation has 3R groups and the right hand side only has one R-group. To fix this we can put a coefficient of 3 in front of CH_3CO_2R on the right hand side giving us the following:

$RCO_2CH_2CH(O_2CR)CH_2CO_2R +CH_3OH \xrightarrow[]{Catalyst} C_3H_8O_3+3CH_3CO_2R$

On the left hand side of the equation we have 7 carbon atoms but on the right hand side we have 9 carbon atoms. This can be rectified by putting a coefficient of 3 in front of CH_3OH on the left hand side of the equation giving us the following:

$RCO_2CH_2CH(O_2CR)CH_2CO_2R +3CH_3OH \xrightarrow[]{Catalyst} C_3H_8O_3+3CH_3CO_2R$

Now all the nonmetals (excluding oxygen) have been balanced and we can move on to step 4

4.) Balance oxygen. There are 9 oxygen atoms on both sides of the equation. Move on to step 5

5.) Balance Hydrogen. There are 17 hydrogens on both sides of the equation.

6.) Recount all atoms to make sure you have balanced correctly. 9 carbons on both sides, 3 R-groups on both sides, 9 oxygen on both sides, and 17 hydrogens on both sides.

If this problem had involved ionic species you would also want to make certain that the charges are balanced on both sides of the equation.

The charges on both sides add up to zero so your equation is balanced:

$RCO_2CH_2CH(O_2CR)CH_2CO_2R +3CH_3OH \xrightarrow[]{Catalyst} C_3H_8O_3+3CH_3CO_2R$

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AP Chemistry : AP Chemistry

Study concepts, example questions & explanations for AP Chemistry

All AP Chemistry Resources

Example Questions

Example Question #3 : Balancing Chemical Equations

Example Question #1 : Balancing Equations

Example Question #2 : Balancing Equations

Example Question #51 : Chemical Reactions

Example Question #52 : Chemical Reactions

Example Question #1 : Balancing Chemical Equations

Example Question #53 : Chemical Reactions

Example Question #54 : Chemical Reactions

Example Question #1 : Balancing Equations

Example Question #11 : Stoichiometry

Transesterification is an industrially important process for the production of biodiesel fuel (that most diesel engines can run cleanly on) and glycerin from triglycerides (usually people use vegetable oil as their feedstock).

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