A) To determine the limiting reagent, we need to compare the number of moles of each reactant to the stoichiometric ratio in the balanced equation. The balanced equation is:
Fe + O2 -> Fe2O3
Calculate the number of moles of iron (Fe):
moles of Fe = mass / molar mass = 72.0 g / 55.845 g/mol ≈ 1.29 mol
Calculate the number of moles of oxygen (O2):
moles of O2 = mass / molar mass = 77.0 g / 32.00 g/mol ≈ 2.41 mol
Now, let's compare the moles of each reactant to the stoichiometric ratio in the balanced equation:
From the balanced equation, 1 mol of Fe reacts with 1 mol of O2 to produce 1 mol of Fe2O3.
The ratio of Fe to O2 is 1.29 mol : 2.41 mol ≈ 0.54.
Since the stoichiometric ratio is 1:1, oxygen is the limiting reagent because you need 1 mol of oxygen for every 1 mol of iron.
B) To find out how much iron(III) oxide is produced, we'll use the balanced equation to determine the number of moles of Fe2O3 and then convert that to grams:
Moles of Fe2O3 that can be produced = moles of limiting reagent (oxygen) = 2.41 mol
Using the molar mass of Fe2O3:
molar mass of Fe2O3 = 2 * atomic mass of Fe + 3 * atomic mass of O = 2 * 55.845 g/mol + 3 * 16.00 g/mol ≈ 159.69 g/mol
Calculate the mass of Fe2O3:
mass = moles * molar mass = 2.41 mol * 159.69 g/mol ≈ 384.78 g
So, the mass of iron(III) oxide produced is approximately 384.78 grams.