1 304 g of aluminum reacts with 742 g of iodine to form alum

1. 3.04 g of aluminum reacts with 7.42 g of iodine to form aluminum iodide. hen the reaction is complete, how many grams of each species are predicted to be present in the reaction flask? b. If the reaction produced 7.25 g of aluminum iodide, what was the percent yield? 2. 45.69 g of solid phosphonus (P)reacts with 131.3 g of chlorine gas to form liquid phosphorus trichloride When the reaction is complete, how many grams of each species are predicted to be present in the reaction flask? a. b. If the reaction produced 145.7 g in the lab, what was the percent yield?

Solution

1)

2 Al (s)   + 3 I2    --------------> 2 AlI3

53.96 g      761.43 g                 815.39 g

3.04 g       7.42 g

53.96 g Al   ---------------> 761.43 g I2

3.04 g Al   --------------> ??

mass of I2 needed = 3.04 x 761.43 / 53.96 = 42.9 g

but here we have only 7.42 g . so

limiting reagent is iodine

mass of AlI3 formed = 7.42 x 815.39 / 761.43

mass of AlI3 formed = 7.95 g

mass of Al = 2.51 g

mass of I2 = 0.00 g

b)

percent yield = actual yield / theoretical ) x 100

                    = 7.25 / 7.95 ) x 100

                     = 91.2 %