Answer
The percent yield of $B_2H_6$ is equal to 67.1%
Work Step by Step
1. Calculate the number of moles of $NaBH_4$:
22.99* 1 + 10.81* 1 + 1.008* 4 = 37.832g/mol
$1.203g \times \frac{1 mol}{ 37.832g} = 0.0318mol (NaBH_4)$
- The balanced reaction is:
$2 NaBH_4 + I_2 -- \gt B_2H_6 + 2NaI + H_2$
The ratio of $NaBH_4$ to $B_2H_6$ is 2 to 1:
$0.0318 mol (NaBH_4) \times \frac{ 1 mol(B_2H_6)}{ 2 mol (NaBH_4)} = 0.0159mol (B_2H_6)$
2. Calculate the mass of $B_2H_6$:
10.81* 2 + 1.008* 6 = 27.668g/mol
$0.0159 mol \times \frac{ 27.668 g}{ 1 mol} = 0.4399g (B_2H_6)$
3. Now, calculate the percent yield:
$\frac{ 0.295g}{0.4399g} \times 100\% = 67.1\%$
