Question

in a reactor0.150 l of oxygen at stp reacts with 32.5 ml of 0.547 m hydrochloric acid. If 0.530 g of chlorine gas forms at stp in this reaction what is the percent yield  

Answer 1

4 HCl (aq) + O2 (g) → 2 Cl2 (g) + 2H2O (g)

At STP , no of mole of O2 gas = PV/RT = 1*0.15/(0.0821*273.15) = 0.0067 mole

At STP , no of mole of HCl gas = 32.5/1000*0.547 = 0.178 mole

limiting reactant = O2

theoretical yield of Cl2 = 2*0.0067 = 0.134 mole

At STP , no of mole of Cl2 gas produced ( practical yield) = 0.53/71 = 0.0075 mole

percent yield=  practical yield /theoretical yield *100

                          = 0.0075/0.134*100

                       = 5.6%