Limiting reagents

Question

Fe₂O₃ + 3CO = 2Fe +3CO₂

If you react 2829g of Fe₂O₃ with unlimited CO. How many grams of Fe do you obtain ?

Answer 1

In order to answer this question, you first need to know the molar mass of the Fe₂O₃. You can calculate this by adding up the atomic masses of each of the atoms in the molecule. So for instance, Fe has an atomic mass of 55.847 amu/mol. Since there are two Fe atoms in each molecule of Fe₂O₃, we need to add 55.847 amu/mol twice. We then add one atom of C (atomic mass 12.001 amu/mol) and three atoms of O (atomic mass 15.999 amu/mol) which gives us a total molecular mass (since were talking about a molecule now, not atoms) of 159.69 amu/mol. This also means that 159.69g of Fe₂O₃ is equal to one mole of Fe₂O₃.

Now that we have the molar mass of Fe₂O₃, we have to determine how many moles of Fe₂O₃ we have. We do this by dividing the given mass of Fe₂O₃ by the molar mass we found in the first part, above. Once we divide 2829g Fe₂O₃ by 159.69 g/mol Fe₂O₃, we find that we have 17.716mol of Fe₂O₃.

Now we need to really look at the chemical reaction. More specifically, we need to look at the coefficients in front of each of the reactants and products. Those coefficients tell us how much of each reactant is needed in order to produce an amount of product. In this problem, there is no coefficient in front of Fe₂O₃, so we can assume its coefficient is 1. The 2 in front of Fe tells us that every time this reaction is run, 2 moles of Fe will be produced. SO... for every 1 mole of Fe₂O₃ we put into the reaction, 2 moles of Fe will be produced.

If we put 17.716 mol of Fe₂O₃, which we found above, all we need to do is double that number to get our final answer: 35.52 mol Fe.