Question

Consider the reaction between HCl and O2: 4HCl(g)+O2(g)→2H2O(l)+2Cl2(g) When 63.1 g of HCl are allowed to react with 17.2 g of O2, 53.8 g of Cl2 are collected. Determine the theoretical yield of Cl2 and the percent yield for the reaction.

Answer 1

first lets convert all given mass data to moles as below and find the limiting reagent and expected or theroretical yields     

Moles of HCl = 63.1/36.5 = 1.728 moles    limiting agent as per stiochiometry

Moles of O2 = 17.2/32 = 0.5375

Moles of Cl₂actually formed = 53.8/71 = 0.7577 moles

Expected moles of Cl₂   are calculated as below :

               4HCl_(g)      +      O_2(g)     →     2H₂O_(l)     +         2Cl_2(g)

          1.728 moles

The moles of Cl₂ = 1/2 moles of HCl consumed =1/2 ( 1.728 moles ) = 0.864 moles

theoretical moles or expected = 0.864 moles

% yield = (moles actually obtained / Theoretical moles ) x100 = (0.7577/0.864)x100 = 87.6 %