Diborane (B2H6, molar mass 27.69 g/mol) can be made from sodium borohydride (NaBH, molar mass 37.85 g/mol) by reacting it with iodine in the following reaction: 2NaBH4 (s)+12 B2He (g)+ 2 Nal (s)+ H2 (9) If 1.203 g NaBH4 is combined with excess l2 and 0.295 g B2Hs is isolated, what is the percent yield of B2H6?
The Correct Answer and Explanation is :
The percent yield of B2H6 (diborane) is approximately 67.04%.
Explanation:
The reaction provided involves the conversion of sodium borohydride (NaBH4) into diborane (B2H6) through a reaction with iodine (I2). To calculate the percent yield, we follow these steps:
- Convert mass of NaBH4 to moles:
The molar mass of NaBH4 is 37.85 g/mol. Given that 1.203 grams of NaBH4 is used, we calculate the moles of NaBH4: [
\text{moles of NaBH4} = \frac{1.203 \, \text{g}}{37.85 \, \text{g/mol}} \approx 0.0318 \, \text{mol}
] - Determine moles of B2H6 produced:
According to the balanced chemical equation, 2 moles of NaBH4 produce 1 mole of B2H6. Thus, the moles of B2H6 produced theoretically can be calculated as: [
\text{moles of B2H6 (theoretical)} = \frac{0.0318 \, \text{mol NaBH4}}{2} = 0.0159 \, \text{mol B2H6}
] - Convert mass of B2H6 obtained to moles:
The molar mass of B2H6 is 27.69 g/mol. With 0.295 grams of B2H6 isolated, we convert this to moles: [
\text{moles of B2H6 (obtained)} = \frac{0.295 \, \text{g}}{27.69 \, \text{g/mol}} \approx 0.0106 \, \text{mol B2H6}
] - Calculate percent yield:
The percent yield is the ratio of the actual yield (moles obtained) to the theoretical yield (moles predicted) multiplied by 100: [
\text{percent yield} = \left( \frac{0.0106 \, \text{mol B2H6}}{0.0159 \, \text{mol B2H6}} \right) \times 100 \approx 67.04\%
]
This means that 67.04% of the theoretical B2H6 was produced in the reaction.
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