Consider the balanced chemical reaction when phosphorus and iodine react to produce phosphorus triodide: 2 P (s) + 3 I2 (g) → 2 PI3 (g) If 58.6 g of I2 (s) area reacted with excess P (s) to produce 0.147 mol of PI3 (g), what is the percent yield of PI3 (g) ?

Question

Deleon

Chemistry

17 August, 19:00

Consider the balanced chemical reaction when phosphorus and iodine react to produce phosphorus triodide: 2 P (s) + 3 I2 (g) → 2 PI3 (g) If 58.6 g of I2 (s) area reacted with excess P (s) to produce 0.147 mol of PI3 (g), what is the percent yield of PI3 (g) ?

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Answers (1)

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Adriana Esparza · Answer 1

Percent yield of PI3 = 95.4%

Explanation:

This is the reaction:

2P (s) + 3I2 (g) > 2PI3 (g)

Let's determine the moles of iodine that has reacted.

58.6 g / 253.8 g/mol = 0.231 mol

Ratio is 3:2. Let's make a rule of three to state the moles produced at 100 % yield reaction.

3 moles of I2 react to make 2 moles of PI3

0.231 moles of I2 would make (0.231.2) / 3 = 0.154 moles of PI3

As we have produced 0.147 moles let's determine the percent yield.

(Yield produced / Theoretical yield). 100 > (0.147 / 0.154). 100 = 95.4%