This question assesses the skill of stoichiometry. The balanced equation CuO + CO → Cu + CO2 provides mole ratios based on the coefficients, which are all 1 in this case. These ratios indicate that 1 mole of CuO produces 1 mole of CO2. To find the moles of CO2 from 2.0 mol of CuO with excess CO, apply the 1:1 ratio directly, resulting in 2.0 mol of CO2. A tempting distractor is 4.0 mol, which might arise from mistakenly doubling the amount due to confusion with a different equation's coefficients. Always start from the balanced equation and convert to moles before applying ratios.

This question assesses the skill of stoichiometry. The balanced equation 2Al + 3Br2 → 2AlBr3 provides mole ratios, with 2 moles of Al producing 2 moles of AlBr3, or a 1:1 ratio. These ratios relate the given moles of Al to the moles of AlBr3. For 0.300 mol of Al with excess Br2, the amount is 0.300 mol of AlBr3. A tempting distractor is 0.450 mol, possibly from mistakenly using the Br2 coefficient in the ratio. Always start from the balanced equation and convert to moles before applying ratios.

This question assesses the skill of stoichiometry. The balanced equation Zn + 2HCl → ZnCl2 + H2 provides mole ratios, with 2 moles of HCl producing 1 mole of H2. These ratios link the given moles of HCl to the moles of H2. For 0.500 mol of HCl with excess Zn, divide by 2 to obtain 0.250 mol of H2. A tempting distractor is 0.500 mol, perhaps from assuming a 1:1 ratio without the coefficient. Always start from the balanced equation and convert to moles before applying ratios.

This question tests stoichiometry. The balanced equation N₂(g) + 3H₂(g) → 2NH₃(g) provides mole ratios, such as 1 mol N₂ to 2 mol NH₃. These ratios allow us to convert the given moles of N₂ to moles of NH₃. For 0.80 mol N₂, the calculation is 0.80 mol N₂ × (2 mol NH₃ / 1 mol N₂) = 1.6 mol NH₃. A tempting distractor is 0.80 mol, which results from mistakenly using a 1:1 ratio instead of 2:1. Always start from the balanced equation and convert to moles before applying ratios.

This stoichiometry problem requires finding aluminum chloride produced from chlorine gas. The balanced equation shows that $3 \text{ mol } \text{Cl}_2$ produces $2 \text{ mol } \text{AlCl}_3$, establishing a 3:2 ratio. Starting with 0.90 mol Cl₂: $(0.90 \text{ mol } \text{Cl}_2) \times \left( \frac{2 \text{ mol } \text{AlCl}_3}{3 \text{ mol } \text{Cl}_2} \right) = 0.60 \text{ mol } \text{AlCl}_3$. Choice C ($0.90 \text{ mol}$) incorrectly assumes equal moles of reactant and product. Always apply the stoichiometric coefficients as a ratio to convert between different substances in the reaction.

This problem applies stoichiometry to find moles of HCl consumed in a reaction with calcium carbonate. The balanced equation CaCO₃ + 2HCl → CaCl₂ + CO₂ + H₂O indicates that 1 mole of CaCO₃ reacts with 2 moles of HCl, establishing a 2:1 ratio. Starting with 0.30 mol CaCO₃, we calculate: 0.30 mol CaCO₃ × (2 mol HCl/1 mol CaCO₃) = 0.60 mol HCl. A common mistake would be assuming a 1:1 ratio and answering 0.30 mol, ignoring the coefficient of 2 for HCl. Always check the coefficients in the balanced equation carefully, as they directly determine the mole ratios needed for calculations.

This problem uses stoichiometry to calculate magnesium nitride formation from magnesium metal. The balanced equation 3Mg + N₂ → Mg₃N₂ reveals that 3 moles of Mg produce 1 mole of Mg₃N₂, establishing a 1:3 ratio. From 0.90 mol Mg, we calculate: 0.90 mol Mg × (1 mol Mg₃N₂/3 mol Mg) = 0.30 mol Mg₃N₂. A common error would be to multiply by 3 instead of divide, giving 2.7 mol, which reverses the relationship between reactant and product. Always verify your mole ratio by checking that units cancel properly: mol Mg × (mol Mg₃N₂/mol Mg) = mol Mg₃N₂.

This problem requires stoichiometry to determine the moles of CO₂ produced from propane combustion. The balanced equation C₃H₈ + 5O₂ → 3CO₂ + 4H₂O shows that 1 mole of C₃H₈ produces 3 moles of CO₂, giving a mole ratio of 3:1. To find moles of CO₂ from 2.0 mol C₃H₈, multiply: 2.0 mol C₃H₈ × (3 mol CO₂/1 mol C₃H₈) = 6.0 mol CO₂. A common error would be to use the coefficient of O₂ (5) instead of CO₂ (3), which would incorrectly give 10 mol as the answer. Always identify the correct mole ratio from the balanced equation by finding the coefficients of your given and desired substances.

This problem requires stoichiometry to find how much oxygen is consumed when magnesium reacts. The balanced equation 2Mg(s) + O₂(g) → 2MgO(s) provides the mole ratio: 2 moles of Mg react with 1 mole of O₂. Starting with 1.00 mol Mg, we calculate: 1.00 mol Mg × (1 mol O₂)/(2 mol Mg) = 0.50 mol O₂. A common error would be to use the coefficient 2 incorrectly, perhaps thinking that 1.00 mol Mg requires 2.00 mol O₂ (answer D), which reverses the actual ratio. Always write out the mole ratio as a fraction with the desired substance in the numerator and the given substance in the denominator, using coefficients from the balanced equation.

This problem involves stoichiometry in a decomposition reaction to find moles of oxygen produced. The balanced equation 2H₂O₂(aq) → 2H₂O(l) + O₂(g) shows that 2 moles of H₂O₂ produce 1 mole of O₂. Starting with 0.80 mol H₂O₂, we calculate: 0.80 mol H₂O₂ × (1 mol O₂)/(2 mol H₂O₂) = 0.40 mol O₂. A common error would be to use a 1:1 ratio, thinking each H₂O₂ produces one O₂, giving 0.80 mol (answer C), but the balanced equation clearly shows the 2:1 ratio. Always use the coefficients from the balanced equation to establish the correct mole ratio, even when it seems counterintuitive.