For this question, we have the following reaction:
Fe2O3 + 3 CO -> 2 Fe + 3 CO2
For letter D, we have:
10 grams of Fe2O3, molar mass = 159.7g/mol
6 grams of CO, molar mass = 28g/mol
In this case, we need to find which compound is the limiting and which is the excess reactant, and it is important to have in mind the molar ratio concept, which we use the coefficients in front of the compound to determine the ratios, like for Fe2O3 and CO, the molar ratio is 1:3. Now let's check if Fe2O3 is in excess, finding the number of moles first:
159.7g = 1 mol
10g = x moles
x = 0.063 moles of Fe2O3
According to the molar ratio, we need 3 times this value for CO, therefore, 0.063 * 3 = 0.189 moles of CO, now we need to check if this is the amount that we have of CO, or if we have more than that:
28g = 1 mol
6g = x moles
x = 0.214 moles of CO, this means that we have more CO than we actually need, making it the excess reactant and Fe2O3 the limiting reactant
Now to find the mass of Fe, we will be using the number of moles of the reactant, which is Fe2O3, 0.063 moles, and according to the molar ratio, we have twice this value for Fe, therefore, 0.063 * 2 = 0.126 moles of Fe, now using the molar mass of Fe, 55.84g/mol, we can find the final mass:
55.84g = 1 mol
x grams = 0.126 moles of Fe
x = 7.03 grams of Fe are produced, or if it is possible to round up, 7 grams of Fe, this is letter D
E. In this case, we need to see how much CO remains in the reaction, and since we need only 0.189 moles of CO and we actually have 0.214 moles, we have a 0.025 moles leftover, which is, in grams:
28g = 1 mol
x grams = 0.025 moles
x = 0.7 grams of CO remaining, only 5.3 grams will react
F. The percent yield is calculated having the actual yield, given in the question and the theoretical yield, which is the value we found in letter D, the formula for it is:
%yield = (actual yield/theoretical yield)*100
%yield = (6.75g/7.03g)*100
%yield = 0.964*100
The percent yield will be 96.4% for this reaction