Question

Sodium borohydride $\mathbf{\left(}{\mathbf{NaBH}}_{\mathbf{4}}\mathbf{\right)}$ is used industrially in many organic syntheses. One way to prepare it is by reacting sodium hydride with gaseous diborane $\left(B_2{H}_6\right)$Assuming an 88.5% yield, how many grams of ${\mathbf{NaBH}}_{\mathbf{4}}$can be prepared by reacting 7.98 g of sodium hydride and 8.16 g of diborane?

Short answer

Mass of ${\mathbf{NaBH}}_{\mathbf{4}}$ formed is 11.15 g.

Step-by-step solution

Writing balanced equation

NaBH₄ is prepared using the following equation:

${\mathbf{B}}_{\mathbf{2}}\mathbf{H}\mathbf{+}\mathbf{2}\mathbf{NaH}\mathbf{\to }\mathbf{2}{\mathbf{NaBH}}_{\mathbf{4}}$

2 moles of NaH and 1 mole of diborane react together to give 2 moles of ${\mathbf{NaBH}}_{\mathbf{4}}$

Calculating moles of reactants

Moles of NaH can be calculated as:

$\mathbf{Mole}\mathbf{}\mathbf{of}\mathbf{}\mathbf{NaH}\mathbf{=}\frac{\mathbf{Mass}\mathbf{}\mathbf{of}\mathbf{}\mathbf{NaH}}{\mathbf{Molar}\mathbf{}\mathbf{Mass}}\mathbf{=}\frac{\mathbf{7}\mathbf{.}\mathbf{98}}{\mathbf{24}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{3325}\mathbf{mole}$

Moles of${\mathbf{B}}_{\mathbf{2}}{\mathbf{H}}_{\mathbf{6}}$can be calculated as:

$\mathbf{Mole}\mathbf{}\mathbf{of}\mathbf{}{\mathbf{B}}_{\mathbf{2}}{\mathbf{H}}_{\mathbf{6}}\mathbf{=}\frac{\mathbf{Mass}\mathbf{}\mathbf{of}\mathbf{}\mathbf{}{\mathbf{B}}_{\mathbf{2}}{\mathbf{H}}_{\mathbf{6}}}{\mathbf{Molar}\mathbf{}\mathbf{Mass}}\mathbf{=}\frac{\mathbf{8}\mathbf{.}\mathbf{16}}{\mathbf{27}\mathbf{.}\mathbf{66}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{295}\mathbf{mole}$

From the balanced equation we can say that NaH is limiting reagent

Moles of NaBH4  formed

From the balanced equation it is clear that 2 moles of NaH gives 2 moles ofso 0.3325 mole of${\mathbf{NaBH}}_{\mathbf{4}}$ NaH will give = 0.3325 mole of${\mathbf{NaBH}}_{\mathbf{4}}$

Mass of ${\mathbf{NaBH}}_{\mathbf{4}}$ formed if yield is 100% (theoretical yield) = $\mathbf{37}\mathbf{.}\mathbf{83}\mathbf{\times }\mathbf{0}\mathbf{.}\mathbf{3325}\mathbf{=}\mathbf{12}\mathbf{.}\mathbf{6}\mathbf{g}$

Actual yield

Percent yield =88.5 %

$\mathbf{Actual}\mathbf{}\mathbf{yield}\mathbf{=}\frac{\mathbf{88}\mathbf{.}\mathbf{5}}{\mathbf{100}}\mathbf{\times }\mathbf{theoritical}\mathbf{}\mathbf{yield}\mathbf{=}\frac{\mathbf{88}\mathbf{.}\mathbf{5}}{\mathbf{100}}\mathbf{\times }\mathbf{12}\mathbf{.}\mathbf{6}\mathbf{=}\mathbf{11}\mathbf{.}\mathbf{15}\mathbf{g}$

So, Mass of ${\mathbf{NaBH}}_{\mathbf{4}}$ formed is 11.15 g.