Q91P - Chemistry: Molecular Nature Of Matter And Change

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?

Step-by-Step Solution

Answer

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

Step 1: 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 ${\mathbf{NaBH}}_{\mathbf{4}}$ 

Step 2: 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

Step 3: Moles of NaBH 4 formed

From the balanced equation it is clear that 2 moles of NaH gives 2 moles of   so 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) =  $\mathbf{37}\mathbf{.}\mathbf{83}\mathbf{\times }\mathbf{0}\mathbf{.}\mathbf{3325}\mathbf{=}\mathbf{12}\mathbf{.}\mathbf{6}\mathbf{g}$

Step 4: 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.