Question

A chemist wants to prepare diborane by the reaction $6LiH+8B{F}_3\to 6LiB{F}_4+B_2{H}_6$
If he starts with $2.0$ moles each of LiH & $B{F}_3$. How many moles of $B_2{H}_6$ can be prepared?

Answer 1

6 moles of LiH requires 8 moles of $B{F}_3$ to form $B_2{H}_6$

1 moles of LiH requires 8/6 moles of $B{F}_3$

2 moles of LiH requires (8/6) x 2 moles of $B{F}_3$ = 8/3

Hence, $B{F}_3$ is a limiting reagent,

So, Moles produced of $B_2{H}_6$ = (1/8) x 2 = 0.25