The reaction of $$NaBH_4$$ with iodine produces $$BI_3$$ according to the following equation:

$$NaBH_4+4I_2\rightarrow BI_3+NaI+4HI$$

If $$37.8g$$ of $$NaBH_4$$ reacts with excess iodine and produces $$39.18g$$ of $$BI_3$$, calculate the percentage yield of $$BI_3$$.

Given atomic masses: $$Na=23,\ B=10.8,\ I=127$$

Enter the answer as the nearest integer.

1. Calculate Molar Masses

a) Molar mass of $$\mathrm{NaBH_4}$$:

$$23+10.8+(4\times1)=37.8\,\mathrm{g\,mol^{-1}}$$

b) Molar mass of $$\mathrm{BI_3}$$:

$$10.8+(3\times127)=391.8\,\mathrm{g\,mol^{-1}}$$

2. Determine Theoretical Yield

a) Moles of $$\mathrm{NaBH_4}$$ reacted:

$$n=\frac{37.8\,\mathrm{g}}{37.8\,\mathrm{g\,mol^{-1}}}=1\,\mathrm{mol}$$

b) According to the balanced chemical equation, 1 mole of $$\mathrm{NaBH_4}$$ produces 1 mole of $$\mathrm{BI_3}$$.

c) Theoretical mass of $$\mathrm{BI_3}$$ produced:

$$m=1\times391.8=391.8\,\mathrm{g}$$

3. Calculate Percentage Yield

a) Using the percentage yield formula:

$$\mathrm{Percentage\ Yield}=\frac{\mathrm{Actual\ Yield}}{\mathrm{Theoretical\ Yield}}\times100$$

b) Substituting the values:

$$\mathrm{Percentage\ Yield}=\frac{39.18}{391.8}\times100$$

$$\boxed{\mathrm{Percentage\ Yield}=10\%}$$

Correct answer: 10