Question

Consider the following set of reactions:
$Al+6HBr\to AlB{r}_3+H_2{C}_2{H}_2+H_2\to C_2{H}_4$
Find the number of moles of ethylene produced when 81 g of aluminium and 40.5 g of HBr reacts to form hydrogen which in turn reacts with 1 g of $C_2{H}_2$ to form $C_2{H}_4$.
(Molar mass of Al = 27 g/mol)

• A. 0.056 mol
• B. 0.038 mol
• C. 0.086 mol
• D. 0.018 mol

Answer 1

The correct option is B 0.038 mol

$2Al+6HBr\to 2AlB{r}_3+3H_2{C}_2{H}_2+H_2\to C_2{H}_4$

$\text{Moles of Al =}\frac{\text{given mass}}{\text{molar mass}}=\frac{81}{27}=3mol\text{Moles of HBr}=\frac{\text{given mass}}{\text{molar mass}}=\frac{40.5}{81}=0.5mol$
Finding the limiting reagent:
$\text{For Al}=\frac{\text{given moles}}{\text{stoichiometric coefficient}}=\frac{3}{2}=1.5mol\text{For HBr}=\frac{\text{given moles}}{\text{stoichiometric coefficient}}=\frac{0.5}{6}=0.083mol$
So, HBr is the limiting reagent.
6 mol of HBr produce 3 mol of $H_2$
0.5 mol will produce = $\frac{0.5}{2}=0.25molof{H}_2$
In reaction (ii),
Moles of $C_2{H}_2=\frac{1}{26}=0.038$ mol
So, the limiting reagent will be ethylene.
1 mol of $C_2{H}_2$ produces 1 mol of $C_2{H}_4$
0.038 mol of $C_2{H}_2$ will produce 0.038 mol of $C_2{H}_4$.