Question

A sample of 10.1 g of Fe2O3 reacts with 12.8 g CO to yield Fe and CO2. The balanced chemical equation is Fe2O3(s)+3CO(g)⟶2Fe(s)+3CO2(g) Which substance is the limiting reactant? CO Fe CO2 Fe2O3 What is the theoretical yield of Fe ? mass of Fe: g If the actual experimental yield for Fe is 6.68 g , what is the percent yield of Fe ? percent yield:

Answer 1

The limiting reactant is CO, and the theoretical yield of Fe is 1.76 g. The percent yield of Fe is 379.55%.

Step-by-step explanation:

The first step in determining the limiting reactant is to calculate the moles of each reactant. To do this, we use the molar mass of Fe2O3 and CO and the given masses of each substance. The molar mass of Fe2O3 is 159.70 g/mol, so the number of moles of Fe2O3 is 10.1 g / 159.70 g/mol = 0.0633 mol. The molar mass of CO is 28.01 g/mol, so the number of moles of CO is 12.8 g / 28.01 g/mol = 0.456 mol.

Next, we use the stoichiometric coefficients from the balanced chemical equation to determine the mole ratio between Fe2O3 and CO. The ratio is 1:3, so for every 1 mol of Fe2O3, we need 3 mol of CO.

Since the ratio of Fe2O3 to CO is 1:3, and we have 0.0633 mol of Fe2O3 and 0.456 mol of CO, we can see that the CO is the limiting reactant because we have less of it compared to the stoichiometric ratio. Therefore, the correct answer is CO.

The theoretical yield of Fe can be calculated using the mole ratio between Fe2O3 and Fe. From the balanced chemical equation, we can see that the ratio is 2:1, so for every 2 mol of Fe2O3, we get 1 mol of Fe. Since we have 0.0633 mol of Fe2O3, the theoretical yield of Fe is 0.0633 mol of Fe * (1 mol of Fe / 2 mol of Fe2O3) * (55.85 g/mol of Fe) = 1.76 g of Fe.

The percent yield of Fe can be calculated by dividing the actual experimental yield (6.68 g) by the theoretical yield (1.76 g) and multiplying by 100. So the percent yield of Fe is (6.68 g / 1.76 g) * 100 = 379.55%.