Question

How many moles of I2 are required to react with 0.429 mol of Al according to the following equation ? Al + 3I2 ⟶ 2AlI3

Answer 1

To determine how many moles of \( I_2 \) are required to react with 0.429 moles of \( Al \), we can use the stoichiometry from the balanced equation:

\[
Al + 3I_2 \rightarrow 2AlI_3
\]

From the equation, we see that 1 mole of \( Al \) reacts with 3 moles of \( I_2 \).

Now, we can use this ratio to find the number of moles of \( I_2 \) required for 0.429 moles of \( Al \):

\[
\text{Moles of } I_2 = 0.429 \, \text{moles} \, Al \times \frac{3 \, \text{moles} \, I_2}{1 \, \text{mole} \, Al}
\]

Calculating this gives us:

\[
\text{Moles of } I_2 = 0.429 \times 3 = 1.287 \, \text{moles} \, I_2
\]

Therefore, **1.287 moles of \( I_2 \)** are required to react with 0.429 moles of \( Al \).