Question

part a how many grams of aluminum oxide are produced according to the reaction below given that you start with 10.0 grams of al and 19.0 grams of o₂? reaction: 4al + 3o₂ → 2al₂o₃ 40.4 5.00 18.9 0.185 not enough information submit request answer

Explanation:

Step1: Calculate moles of Al

Molar mass of Al is \( 26.98 \, \text{g/mol} \). Moles of Al: \( \frac{10.0 \, \text{g}}{26.98 \, \text{g/mol}} \approx 0.3706 \, \text{mol} \).

Step2: Calculate moles of \( O_2 \)

Molar mass of \( O_2 \) is \( 32.00 \, \text{g/mol} \). Moles of \( O_2 \): \( \frac{19.0 \, \text{g}}{32.00 \, \text{g/mol}} \approx 0.5938 \, \text{mol} \).

Step3: Determine limiting reactant

From reaction \( 4\text{Al} + 3\text{O}_2
ightarrow 2\text{Al}_2\text{O}_3 \), mole ratio \( \text{Al}:\text{O}_2 = 4:3 \).
For Al: \( \frac{0.3706}{4} = 0.09265 \). For \( O_2 \): \( \frac{0.5938}{3} \approx 0.1979 \). Al is limiting.

Step4: Moles of \( \text{Al}_2\text{O}_3 \) from Al

Mole ratio \( \text{Al}:\text{Al}_2\text{O}_3 = 4:2 = 2:1 \). Moles of \( \text{Al}_2\text{O}_3 \): \( \frac{0.3706}{2} \approx 0.1853 \, \text{mol} \).

Step5: Mass of \( \text{Al}_2\text{O}_3 \)

Molar mass of \( \text{Al}_2\text{O}_3 \): \( 2(26.98) + 3(16.00) = 101.96 \, \text{g/mol} \).
Mass: \( 0.1853 \, \text{mol} \times 101.96 \, \text{g/mol} \approx 18.9 \, \text{g} \).

Answer:

18.9