Question

Diborane $\left( B _{2} H _{6}\right)$ can be prepared by the following reaction-
$3 NaBH _{4}+4 BF _{3} \to 3 NaBF _{4}+2 B _{2} H _{6}$
If the reaction has a $70 \%$ yield, how many moles of $NaBH _{4}$ should be used with excess $BF _{3}$ in order to obtain $0.200 \,mol$ of $B _{2} H _{6}$ ?

• A. 0.21 moles
• B. 0.429 moles
• C. 0.300 mol
• D. 0.175 moles

Answer 1

Answer: (B)

$3 \,NaBH _{4}+4 BF _{3} \to 3 NaBF _{4}+2 B _{2} H _{6}$

Since $BF _{3}$ is in excess, the limiting reagent is $NaBH _{4}$.

To obtain 2 mole $B _{2} H _{6}, NaBH _{4}$ required $=3$ mole

To obtain 0.200 mole $B_{2} H_{6},$ $NaBH _{4}$ required $=\frac{3}{2} \times 0.200$

$=0.300 \,mole$

Because, the yield is $70 \%,$ Hence

$x \times \frac{70}{100}=0.300$

$x=\frac{0.300}{1000} \times \frac{100}{70}=\frac{3}{7}=0.429$ moles