Question

Consider the reaction between HCl and O2:
4HCl(g)+O2(g)→2H2O(l)+2Cl2(g) When 63.1 g of HClare allowed to react with 17.2 g of O2, 51.9 g of Cl2 are collected.

Determine the limiting reactant for the reaction.

Determine the percent yield for the reaction.

Answer 1

Number of moles of HCl = 63.1 g / 36.46 g/mol = 1.73 mole

Number of moles of O2 = 17.2 g / 32.0 g/mol = 0.538 mole

From the balanced equation we can say that

4 mole of HCl requires 1 mole of O2 so

1.73 mole of HCl will require

= 1.73 mole of HCl *(1 mole of O2 / 4 mole of HCl)

= 0.433 mole of O2

But we have 0.538 mole of O2 which is in excess so O2 is an excess reactant and

HCl is the limiing reactant

From the balanced equation we can say that

4 mole of HCl produces 2 mole of Cl2 so

1.73 mole of HCl will produce

= 1.73 mole of HCl *( 2 mole of Cl2 / 4 mole of HCl)

= 0.865 mole of Cl2

mass of 1 mole of Cl2 = 70.906 g so

the mass of 0.865 mole of Cl2 = 61.3 g

Therefore, theoretical yield of Cl2 = 61.3 g

percent yield = (actual yield / theoretical yield)*100

percent yield = (51.9 / 61.3)*100 = 84.7 %

Therefore, percent yield = 84.7 %