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 H2 + Cl2 → 2HCl provides mole ratios, with 1 mole of H2 producing 2 moles of HCl. These ratios link the given moles of H2 to the moles of HCl. With 0.150 mol of H2 and excess Cl2, multiply by 2 to get 0.300 mol of HCl. A tempting distractor is 0.150 mol, which could result from forgetting to apply the coefficient and using a 1:1 ratio. 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 assesses the skill of stoichiometry. The balanced equation CaCO3 → CaO + CO2 provides mole ratios where 1 mole of CaCO3 produces 1 mole of CO2. These ratios allow conversion from grams of CaCO3 to moles using its molar mass, then to moles of CO2, and finally to grams of CO2 using its molar mass. For 10.0 g of CaCO3 (0.100 mol), the 1:1 ratio yields 0.100 mol of CO2, or 4.4 g. A tempting distractor is 44 g, which could result from forgetting to convert grams to moles and directly using the molar mass without 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 N2 + 3H2 → 2NH3 provides mole ratios, including 3 moles of H2 to 2 moles of NH3. These ratios connect the given moles of H2 to the moles of NH3. With 0.900 mol of H2 and excess N2, multiply by 2/3 to get 0.600 mol of NH3. A tempting distractor is 0.900 mol, which might arise from assuming a 1:1 ratio without considering coefficients. 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 question tests stoichiometry. The balanced equation 2NO₂(g) → N₂O₄(g) provides mole ratios, such as 2 mol NO₂ to 1 mol N₂O₄. These ratios relate the given moles of NO₂ to moles of N₂O₄. For 1.2 mol NO₂, the calculation is 1.2 mol NO₂ × (1 mol N₂O₄ / 2 mol NO₂) = 0.60 mol N₂O₄. A tempting distractor is 1.2 mol, resulting from assuming a 1:1 ratio without considering the coefficient 2. Always start from the balanced equation and convert to moles before applying ratios.

This problem requires stoichiometry to calculate moles of Fe produced from CO. The balanced equation Fe₂O₃(s) + 3CO(g) → 2Fe(s) + 3CO₂(g) shows that 3 moles of CO produce 2 moles of Fe, giving a 3:2 ratio. With 0.90 mol CO and excess Fe₂O₃, we calculate: 0.90 mol CO × (2 mol Fe/3 mol CO) = 0.60 mol Fe. A common mistake would be to use a 1:1 ratio or to multiply by 2 without considering the coefficient 3 for CO, yielding 1.80 mol (answer B). The key strategy is to set up the mole ratio using the exact coefficients from the balanced equation, ensuring the given substance is in the denominator and the desired product is in the numerator.

This problem requires stoichiometry to find the moles of water produced from oxygen gas. The balanced equation shows that 1 mol O₂ produces 2 mol H₂O, giving a mole ratio of 2:1. Starting with 3.0 mol O₂, we multiply by the ratio (2 mol H₂O/1 mol O₂) to get 6.0 mol H₂O. A common mistake would be to use 3.0 mol (choice B), forgetting to apply the stoichiometric coefficient. Always start from the balanced equation and use the coefficients as mole ratios to convert between reactants and products.