Question

If 48.0 g of iodine are reacted with an excess of phosphorus according to the following chemical equation:

2 P(s) + 3 I2(s) → 2 PI3(s)

Calculate the theoretical yield of PI3.

If the actual yield of PI3 is 50.0 g, what is the percentage yield?

Answer 1

Theoretical Yield = 51.9 grams PI₃

Percentage Yield = 96.3% PI₃

Step-by-step explanation:

To find the theoretical and percent yield of PI₃, you need to (1) convert grams I₂ to moles I₂ (via molar mass), then (2) convert moles I₂ to moles PI₃ (via mole-to-mole ratio from reaction coefficients), then (3) convert moles PI₃ to grams PI₃ (via molar mass), and then (4) calculate the percent yield. It is important to arrange the ratios/conversions in a way that allows for the cancellation of units. The final answers should have 3 sig figs to match the sig figs of the given values.

Molar Mass (I₂): 2(126.90 g/mol)

Molar Mass (I₂): 253.8 g/mol

Molar Mass (PI₃): 30.974 g/mol + 3(126.90 g/mol)

Molar Mass (PI₃): 411.674 g/mol

2 P(s) + 3 I₂(s) -----> 2 PI₃(s)

48.0 g I₂ 1 mole 2 moles PI₃ 411.674 g
--------------- x ----------------- x ---------------------- x ------------------ =
253.8 g 3 moles I₂ 1 mole

= 51.9 g PI₃

Actual Yield
Percentage Yield = --------------------------------- x 100%
Theoretical Yield

50.0 g PI₃
Percentage Yield = -------------------- x 100%
51.9 g PI₃

Percentage Yield = 96.3% PI₃