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Question

The reaction between NaBH₄ and I₂ will form B₂H₆, NaI, and H₂. Suppose you use 1.203 g of NaBH₄ with an excess of iodine and experimentally obtain 0.295 g of B₂H₆. What is the percent yield of your work?

1. a. 150%
2. b. 67.1%
3. :c. 32.9%
4. d. 49.1%

Answer 1

The first step in solving problems of this sort is to write a balanced chemical equation.

In this case,

2 NaBH4 + I2 --->. B2H6 + 2 NaI + H2

All chemical equations express quantities of materials as MOLES, not GRAMS. In this case,

two moles of sodium borohydride react with one mole of iodine to yield one mole of diborane, two moles of sodium iodide, and one mole of hydrogen gas.

The next step is to convert any quantities expressed as grams into moles. You do that by dividing the number of grams by the molecular weight, which has the units of (grams/mole), to get moles.

In this case, the molecular weight of sodium borohydride is 37.83 grams per mole and the molecular weight of diborane is 27.67 grams per mole.

From the problem, you start with 1.203 g of sodium borohydride. 1.203/37.83 = 0.03180 moles of sodium borohydride.

You wind up with 0.295 grams of diborane. 0.295/27.67 = 0.01066 moles of diborane.

Now, look at the chemical equation. If ALL of the sodium borohydride had been converted to diborane, and you started with 0.0318 moles of sodium borohydride, you would have wound up with 0.01590 moles of diborane, since for every two moles of sodium borohydride you wind up with one mole of diborane.

Since you actually wound up with 0.0107 moles of diborane, the percent yield is

(0.01066/0.0159)*100% = 67.1%, which is answer b.