Determine which reactant is limiting and calculate the theoretical yield of rust (FeO) formed based on the limiting reactant.
Given reaction: 4Fe + 3O2 → 2Fe2O3 First, we need to find the moles of each reactant using their respective molar masses:
Molar mass of Fe (iron) = 55.85 g/mol
Molar mass of O2 (oxygen) = 32.00 g/mol
1. Moles of iron (Fe):
Moles of Fe = Mass / Molar mass
Moles of Fe = 10.0 g / 55.85 g/mol = 0.179 mol
2. Moles of oxygen (O2):
Moles of O2 = Mass / Molar mass
Moles of O2 = 20.0 g / 32.00 g/mol = 0.625 mol
"The balanced equation tells us that 4 moles of iron react with 3 moles of oxygen. The ratio is 4:3.
Calculate the ratio of actual moles of iron to actual moles of oxygen:
Actual ratio = (moles of Fe) / (moles of O2)
Actual ratio = 0.179 mol / 0.625 mol = 0.2864
Compare this ratio to the ratio from the balanced equation:
Balanced ratio = 4 / 3 = 1.33
The actual ratio (0.2864) is less than the balanced ratio (1.33), which means that iron is the limiting reactant.
Now, calculate the moles of FeO that can be formed based on the limiting reactant (iron):
Moles of FeO = (moles of Fe) * (2 moles FeO / 4 moles Fe)
Moles of FeO = 0.179 mol * (2 / 4) = 0.0895 mol
Finally, calculate the mass of FeO formed using its molar mass:
Molar mass of FeO = 55.85 g/mol + 16.00 g/mol = 71.85 g/mol
Mass of FeO = Moles of FeO * Molar mass of FeO
Mass of FeO = 0.0895 mol * 71.85 g/mol = 6.43 g
Therefore, approximately 6.43 grams of rust (FeO) should form when 10.0 grams of iron reacts with 20.0 grams of oxygen.