Question

A chemist wants to prepare diborane by the following reaction:
$6LiH+{8BF}_3⟶{6LiBF}_4+B_2{H}_6$
He starts with $2.0$ moles each of $LiH$ and ${BF}_3$. How many moles of $B_2{H}_6$ can be prepared?

• A. $0.56$ mol
• B. $0.25$ mol
• C. $0.36$ mol
• D. $0.26$ mol

Answer 1

The correct option is B $0.25$ mol

The balanced reaction is given below:
$6LiH+8B{F}_3\to 6LiB{F}_4+B_2{H}_6$
$2$ $2$ (No. of moles given)
$\frac{2}{6}$ $\frac{2}{8}$ ( Ratio of moles given and stoichiometry coefficient)
As $\frac{2}{8}$ is lower than $\frac{2}{6}$, $B{F}_3$ is limiting reagent and product will form according to number of moles of $B{F}_3$ (given).
So, moles produced of $B_2{H}_6=\frac{1}{8}\times 2=0.25$ mol.