Chapter 3: Q91P (page 135)

Sodium borohydride $({NaBH}_{4})$ is used industrially in many organic syntheses. One way to prepare it is by reacting sodium hydride with gaseous diborane $(B_{2} H_{6})$ Assuming an 88.5% yield, how many grams of ${NaBH}_{4}$ can be prepared by reacting 7.98 g of sodium hydride and 8.16 g of diborane?

Short Answer

Expert verified

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

Step by step solution

Writing balanced equation

NaBH₄ is prepared using the following equation:

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

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

Calculating moles of reactants

Moles of NaH can be calculated as:

$Mole of NaH = \frac{Mass of NaH}{Molar Mass} = \frac{7.98}{24} = 0.3325 mole$

Moles of $B_{2} H_{6}$ can be calculated as:

$Mole of B_{2} H_{6} = \frac{Mass of B_{2} H_{6}}{Molar Mass} = \frac{8.16}{27.66} = 0.295 mole$

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 ${NaBH}_{4}$ NaH will give = 0.3325 mole of ${NaBH}_{4}$

Mass of ${NaBH}_{4}$ formed if yield is 100% (theoretical yield) = $37.83 \times 0.3325 = 12.6 g$

Actual yield

Percent yield =88.5 %

$Actual yield = \frac{88.5}{100} \times theoritical yield = \frac{88.5}{100} \times 12.6 = 11.15 g$

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

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Short Answer

Step by step solution

Writing balanced equation

Calculating moles of reactants

Moles of NaBH4 formed

Actual yield

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Sodium borohydride (NaBH4) is used industrially in many organic syntheses. One way to prepare it is by reacting sodium hydride with gaseous diborane (B2H6)Assuming an 88.5% yield, how many grams of NaBH4can be prepared by reacting 7.98 g of sodium hydride and 8.16 g of diborane?

Short Answer

Step by step solution

Writing balanced equation

Calculating moles of reactants

Moles of NaBH4 formed

Actual yield

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Chemistry Textbooks

Chemical Analysis

Physical Chemistry

Nuclear Chemistry

Organic Chemistry

The Earths Atmosphere

Chemical Reactions

Study anywhere. Anytime. Across all devices.