Answer to Question #131230 in General Chemistry for mike

Question

Phosphorus reacts with iodine as shown in the chemical reaction below. What is the percent yield of the reaction if 28.2 g PI3 is obtained from the reaction of 48.0 g of I2 with excess phosphorus?
2P(s) + 3I2(s) ® 2PI3(s)

Answer 1

Molar mass of I₂ = 126.9×2 g/mol = 253.8 g/mol

Mol of I₂ in 48.0 g = 48.0g/253.8 g/mol = 0.189 mol

From the balanced equation:

3 mol of I₂ produce 2 mol of PI₃

Then, 0.189 mol of I₂ produce = 0.189×2/3 = 0.126 mol PI₃

Molar mass of PI₃ = 411.69 g/mol

Mass of PI₃ produced = 0.126 mol×411.69 g/mol = 51.9 g of PI₃

48.0 g of I₂ produce 51.9 g of PI₃ i.e. theoretical yield of the reaction.  

 Percent of yield = (experimental yield ×100)/theoretical yield

                           = (28.2×100)/51.9

                           = 54.33 (Answer)   

The yield of the reaction is 54.33 percent.