Question

A chemist wants to prepare diborane by the reaction.
$LiH+B{F}_3\to LiB{F}_4+B_2{H}_6$
If he starts with 2.0 moles each of $LiH\&B{F}_3$. Then, the correct option(s) is/are:

• A. A maximum of 0.25 moles of $B_2{H}_6$ can be prepared
• B. More than 0.25 moles of $B_2{H}_6$ can be prepared
• C. Not more than 1.5 moles of $LiB{F}_4$ can be prepared
• D. 2 moles of $LiB{F}_4$ can be prepared

Answer 1

Answer: (A), (C)

The correct options are
A A maximum of 0.25 moles of $B_2{H}_6$ can be prepared
C Not more than 1.5 moles of $LiB{F}_4$ can be prepared

Balanced reaction is
$6LiH+8B{F}_3\to 6LiB{F}_4+B_2{H}_6$
and
Here, $B{F}_3$ is a limiting reagent.
Moles of $B_2{H}_6$ forms from 2 moles of $B{F}_3=\frac{2}{8}=0.25$
Similarly, no. of $LiB{F}_4$ that forms from 2 moles of $B{F}_3$ = $\frac{6}{8}\times 2=1.5$