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2 May, 21:03

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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  1. 2 May, 23:13
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    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%
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