Question

now that we know which reactant is limiting, we can use it to find other information.
4al(s) + 3o₂(g) → 2al₂o₃(s)
molar masses \t26.98 g/mol \t32.00 g/mol \t101.96 g/mol
what mass of o₂ would be used during the reaction with 13.5 g al?

Explanation:

Step1: Calculate moles of Al

Moles of Al = $\frac{\text{mass of Al}}{\text{molar mass of Al}}$ = $\frac{13.5\ \text{g}}{26.98\ \frac{\text{g}}{\text{mol}}}$ ≈ 0.500 mol

Step2: Use stoichiometry to find moles of O₂

From the balanced equation, 4 moles of Al react with 3 moles of O₂. So, moles of O₂ = $\frac{3}{4}$ × moles of Al = $\frac{3}{4}$ × 0.500 mol = 0.375 mol

Step3: Calculate mass of O₂

Mass of O₂ = moles of O₂ × molar mass of O₂ = 0.375 mol × 32.00 $\frac{\text{g}}{\text{mol}}$ = 12.0 g

Answer:

12.0 g