QuestionJuly 28, 2025

Phosphorus reacts with iodine as shown in the chemical reaction below.What is the percent yield of the reaction if 28.2 g Pl_(3) is obtained from the reaction of 48.0 g of 12 with excess phosphorus? 2P(s)+3I_(2)(s)arrow 2PI_(3)(s) 51.7 g 51.48 51.5 g None of the above 51.98

Phosphorus reacts with iodine as shown in the chemical reaction below.What is the percent yield of the reaction if 28.2 g Pl_(3) is obtained from the reaction of 48.0 g of 12 with excess phosphorus? 2P(s)+3I_(2)(s)arrow 2PI_(3)(s) 51.7 g 51.48 51.5 g None of the above 51.98
Phosphorus reacts with iodine as shown in the chemical reaction below.What is the percent
yield of the reaction if 28.2 g Pl_(3) is obtained from the reaction of 48.0 g of 12 with excess
phosphorus?
2P(s)+3I_(2)(s)arrow 2PI_(3)(s)
51.7 g
51.48
51.5 g
None of the above
51.98

Solution
4.5(454 votes)

Answer

None of the above Explanation 1. Calculate moles of I_2 Molar mass of I_2 is 253.8 \, \text{g/mol}. Moles of I_2 = \frac{48.0 \, \text{g}}{253.8 \, \text{g/mol}} = 0.189 \, \text{mol}. 2. Determine moles of PI_3 produced From the balanced equation, 3 \, \text{mol} \, I_2 produces 2 \, \text{mol} \, PI_3. Thus, 0.189 \, \text{mol} \, I_2 produces \frac{2}{3} \times 0.189 = 0.126 \, \text{mol} \, PI_3. 3. Calculate theoretical yield of PI_3 Molar mass of PI_3 is 411.7 \, \text{g/mol}. Theoretical yield = 0.126 \, \text{mol} \times 411.7 \, \text{g/mol} = 51.9 \, \text{g}. 4. Calculate percent yield Percent yield = \left(\frac{28.2 \, \text{g}}{51.9 \, \text{g}}\right) \times 100\% = 54.3\%.

Explanation

1. Calculate moles of $I_2$<br /> Molar mass of $I_2$ is $253.8 \, \text{g/mol}$. Moles of $I_2 = \frac{48.0 \, \text{g}}{253.8 \, \text{g/mol}} = 0.189 \, \text{mol}$.<br />2. Determine moles of $PI_3$ produced<br /> From the balanced equation, $3 \, \text{mol} \, I_2$ produces $2 \, \text{mol} \, PI_3$. Thus, $0.189 \, \text{mol} \, I_2$ produces $\frac{2}{3} \times 0.189 = 0.126 \, \text{mol} \, PI_3$.<br />3. Calculate theoretical yield of $PI_3$<br /> Molar mass of $PI_3$ is $411.7 \, \text{g/mol}$. Theoretical yield $= 0.126 \, \text{mol} \times 411.7 \, \text{g/mol} = 51.9 \, \text{g}$.<br />4. Calculate percent yield<br /> Percent yield $= \left(\frac{28.2 \, \text{g}}{51.9 \, \text{g}}\right) \times 100\% = 54.3\%$.
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