Question 1:

In a chemical reaction, 3 moles of hydrogen gas (H2) react with 2 moles of oxygen gas (O2) to produce water (H2O). What is the limiting reactant if there are 5 moles of hydrogen gas and 4 moles of oxygen gas?

Explanation: To determine the limiting reactant, we compare the moles of each reactant to their respective stoichiometric coefficients. The ratio of moles of hydrogen gas to moles of oxygen gas in the balanced equation is 3:2. For 5 moles of hydrogen gas, the expected moles of oxygen gas required would be (5 * 2) / 3 = 10/3 ≈ 3.33 moles. Since we have only 4 moles of oxygen gas, it is the limiting reactant, option B.

Question 2:

In a reaction, 10 grams of nitrogen gas (N2) react with excess hydrogen gas (H2) to produce ammonia (NH3). If the percent yield of ammonia is 80%, how many grams of ammonia are actually produced?

Explanation: The percent yield is calculated by dividing the actual yield by the theoretical yield and multiplying by 100%. In this case, the theoretical yield of ammonia can be calculated using stoichiometry. The balanced equation shows that 1 mole of nitrogen gas reacts to produce 2 moles of ammonia. From 10 grams of nitrogen gas, we can calculate the theoretical yield of ammonia as (10 g * 2 mol) / (28 g/mol) = 0.714 moles. Since the percent yield is given as 80%, the actual yield of ammonia would be (0.8 * 0.714 mol * 17 g/mol) ≈ 12 grams, option A.

Question 3:

In a chemical reaction, 2 moles of sulfuric acid (H2SO4) react with 3 moles of calcium hydroxide (Ca(OH)2) to produce calcium sulfate (CaSO4) and water (H2O). If 4 moles of sulfuric acid react with excess calcium hydroxide, what is the theoretical yield of calcium sulfate?

Explanation: The stoichiometric ratio between sulfuric acid and calcium sulfate is 2:1. Therefore, if 4 moles of sulfuric acid react, the theoretical yield of calcium sulfate can be calculated as (4 moles * 1 mole) / 2 = 2 moles, option B.

Question 4:

In a reaction, 50 grams of iron (Fe) react with excess oxygen gas (O2) to produce iron(III) oxide (Fe2O3). What is the percent yield of iron(III) oxide if 85 grams of iron(III) oxide are obtained?

Explanation: The theoretical yield of iron(III) oxide can be calculated using stoichiometry. The balanced equation shows that 4 moles of iron react to produce 2 moles of iron(III) oxide. From 50 grams of iron, we can calculate the theoretical yield of iron(III) oxide as (50 g * 2 mol) / (56 g/mol) = 1.79 moles. The percent yield is calculated by dividing the actual yield (85 grams) by the theoretical yield (1.79 moles) and multiplying by 100%. The result is (85 g / 1.79 mol) * 100% ≈ 47.49%, which is approximately 50%, option C.

Question 5:

In a chemical reaction, 2 moles of methane (CH4) react with 4 moles of oxygen gas (O2) to produce carbon dioxide (CO2) and water (H2O). If only 2 moles of oxygen gas are available, what is the limiting reactant?

Explanation: To determine the limiting reactant, we compare the moles of each reactant to their respective stoichiometric coefficients. The ratio of moles of methane to moles of oxygen gas in the balanced equation is 2:4, which simplifies to 1:2. If we have 2 moles of oxygen gas, the expected moles of methane required would be 2 * 1 = 2 moles. Since we have exactly 2 moles of methane available, it is the limiting reactant, option A.