Question

use the balanced equation, $4\ce{al} + 3\ce{o_2} \
ightarrow 2\ce{al_2o_3}$, to answer the following question. how many grams of $\ce{o_2}$ are needed to react completely with 19.0 g of $\ce{al}$? be sure your answer has the correct number of significant figures.

Explanation:

Step1: Calculate moles of Al

Molar mass of Al is 26.98 g/mol. Moles of Al = $\frac{19.0\ g}{26.98\ g/mol} \approx 0.699\ mol$.

Step2: Determine mole ratio of O₂ to Al

From the equation $4\text{Al} + 3\text{O}_2
ightarrow 2\text{Al}_2\text{O}_3$, the mole ratio of $\text{O}_2$ to $\text{Al}$ is $\frac{3}{4}$.

Step3: Calculate moles of O₂

Moles of $\text{O}_2$ = $0.699\ mol\ \text{Al} \times \frac{3\ mol\ \text{O}_2}{4\ mol\ \text{Al}} \approx 0.524\ mol$.

Step4: Calculate mass of O₂

Molar mass of $\text{O}_2$ is 32.00 g/mol. Mass of $\text{O}_2$ = $0.524\ mol \times 32.00\ g/mol \approx 16.8\ g$.

Answer:

16.8