Question

Consider the reaction: 2 Al + 3Br2 → 2 AlBr3 Suppose a reaction vessel initially contains 5.0 mole Al and 6.0 mole Br2. What is in the reaction vessel once the reaction has occurred to the fullest extent possible? chapter 9

Answer 1

Answer:

4 moles of $AlBr_3$ and 1 mole of $Al$ will be present in the reaction vessel

Explanation:

The reaction given in the question is

$2Al + 3 Br_2$ ⇒ $2AlBr_3$

According to the stoichiometric coefficients of the reaction, 2 moles of $Al$ requires 3 moles of $Br_2$ so in this reaction, $Br_2$ is a limiting reagent. So we will consider that $Al$ is in excess.

Now,

Since 3 moles of $Br_2$ requires 2 moles of $Al$

So, for 6 moles of $Br_2$ the moles of $Al$ required = $\frac{2}{3} \times 6$ = 4 moles.

Moles of $Al$ remaining after the completion of reaction = 5 - 4 = 1 mole.

Again,

Since 3 moles of $Br_2$ produces 2 moles of $AlBr_3$

So, moles of $AlBr_3$ produced by 6 moles of $Br_2$ = $\frac{2}{3} \times 6$ = 4 moles.

Therefore, after the completion of reaction, 4 moles of $AlBr_3$ and 1 mole of $Al$ will be present in the reaction vessel.