Question

in limiting reactant problems, we normally begin by calculating the amount of product each reactant can make.
4al(s) + 3o₂(g) → 2al₂o₃(s)
molar masses 26.98 g/mol 32.00 g/mol 101.96 g/mol
13.5 g al reacts with 32.0 g o₂.
what mass of al₂o₃ forms based on 13.5 g al?

Explanation:

Step1: Calculate moles of Al

Moles of Al = mass / molar mass = \( \frac{13.5\ \text{g}}{26.98\ \frac{\text{g}}{\text{mol}}} \)
\( \approx 0.500\ \text{mol} \)

Step2: Use stoichiometry to find moles of \( \text{Al}_2\text{O}_3 \)

From the reaction: \( 4\ \text{mol Al}
ightarrow 2\ \text{mol}\ \text{Al}_2\text{O}_3 \)
Moles of \( \text{Al}_2\text{O}_3 \) from Al = \( 0.500\ \text{mol Al} \times \frac{2\ \text{mol}\ \text{Al}_2\text{O}_3}{4\ \text{mol Al}} \)
\( = 0.250\ \text{mol} \)

Step3: Calculate mass of \( \text{Al}_2\text{O}_3 \)

Mass = moles × molar mass = \( 0.250\ \text{mol} \times 101.96\ \frac{\text{g}}{\text{mol}} \)
\( = 25.49\ \text{g} \approx 25.5\ \text{g} \)

Answer:

The mass of \( \text{Al}_2\text{O}_3 \) formed is approximately \( 25.5\ \text{g} \) (or more precisely \( 25.49\ \text{g} \)).