Question

the unbalanced chemical equation for the decomposition of aluminum oxide is: $\ce{al_{2}o_{3} -> al + o_{2}}$ how many moles of al would be produced from 20 moles of $\ce{al_{2}o_{3}}$? \bigcirc 4 moles \bigcirc 10 moles \bigcirc 20 moles \bigcirc 40 moles

Explanation:

Step1: Balance the chemical equation

The unbalanced equation is \( \text{Al}_2\text{O}_3
ightarrow \text{Al} + \text{O}_2 \). To balance it, we first balance Al and O. The balanced equation is \( 2\text{Al}_2\text{O}_3
ightarrow 4\text{Al} + 3\text{O}_2 \). From this, the mole ratio of \( \text{Al}_2\text{O}_3 \) to \( \text{Al} \) is \( 2:4 \) or simplified \( 1:2 \).

Step2: Use the mole ratio to calculate moles of Al

Given moles of \( \text{Al}_2\text{O}_3 = 20 \) moles. From the mole ratio (1 mole of \( \text{Al}_2\text{O}_3 \) produces 2 moles of Al), we calculate moles of Al as \( 20 \, \text{moles} \times \frac{4}{2} \) (or using the simplified ratio \( 20 \times 2 \))? Wait, no, from the balanced equation \( 2\text{Al}_2\text{O}_3 \) gives \( 4\text{Al} \), so 1 mole of \( \text{Al}_2\text{O}_3 \) gives \( \frac{4}{2} = 2 \) moles of Al. So for 20 moles of \( \text{Al}_2\text{O}_3 \), moles of Al = \( 20 \times 2 = 40 \) moles? Wait, no, wait the balanced equation: \( 2\text{Al}_2\text{O}_3
ightarrow 4\text{Al} + 3\text{O}_2 \). So 2 moles of \( \text{Al}_2\text{O}_3 \) produce 4 moles of Al. So the ratio of \( \text{Al}_2\text{O}_3 \) to Al is \( 2:4 = 1:2 \). So 1 mole of \( \text{Al}_2\text{O}_3 \) produces 2 moles of Al. Therefore, 20 moles of \( \text{Al}_2\text{O}_3 \) will produce \( 20 \times 2 = 40 \) moles? Wait, no, wait the balanced equation: 2 Al₂O₃ gives 4 Al. So 2 moles Al₂O₃ → 4 moles Al. So 1 mole Al₂O₃ → 2 moles Al. So 20 moles Al₂O₃ → 20 * 2 = 40 moles Al.

Answer:

40 moles (corresponding to the option "40 moles")