Question

A mixture of 1 mol of Al and 3 mol of $C{l}_2$ are allowed to react as:
$2Al+3C{l}_2\to 2AlC{l}_3$
Find the mass of excess reagent left.

• A. 220 g
• B. 115 g
• C. 85 g
• D. 106.5 g

Answer 1

The correct option is D 106.5 g

$2Al+3C{l}_2\to 2AlC{l}_3$
Initial number of moles of Al and $C{l}_2$ are 1 mol and 3 mol respectively

For $Al$,
$\frac{\text{Initial number of moles}}{\text{stoichiometric coefficient}}=\frac{1}{2}=0.5$

For $C{l}_2$, $\frac{\text{Initial number of moles}}{\text{stoichiometric coefficient}}=\frac{3}{3}=1$

According to the stoichiometry,
2 mole of $Al$ reacts with 3 mole of $C{l}_2$
1 mole of $Al$ will react with 1.5 mole of $C{l}_2$

Mass of the excess reagent left unreacted = 1.5 $\times$ molar mass of $C{l}_2$
$=1.5\times 71=106.5g$