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, 46.1 g of Cl2 are collected.

Question

A. Determine the limiting reactant for the reaction.
Express your answer as a chemical formula.

 

B. Determine the theoretical yield of Cl2 for the reaction.

 

C. Determine the percent yield for the reaction.

Answer 1

4HCl(g)+O2(g)→2H2O(l)+2Cl2(g)

A.  Moles HCl = 63.1 g x1 mole/36.5 g = 1.73 moles

Moles O2 = 17.2 g x 1 mole/32g = 0.538 moles

Limiting reactant = HCl since the balanced equation indicates 4 moles of HCl are needed per mole of O2 and there aren’t enough. 

B.  Theoretica yield Cl2 = 1.73 mol HCl x 2 mol Cl2/4 mol HCl = 0.865 moles Cl2 x 71g/mol = 61.4 g

C.  Percent yield = 46.1 g/61.4 g (x100%) = 75.1%