Limiting reactant/percent yield problem? Please help.?

Question

If 74.30 g of HCL were produced from 2.13 g oh H2 and an excess of Cl2 according to the reaction:

H2 + Cl2 ------> 2HCl

What was the percent yield of HCl?

I got an answer that is slightly off what the answer key says. If the problem says "excess of Cl2", doesn't that mean H2 is the limiting reactant?

The answer on the answer key was 96.44%.

Answer 1

Yes, it means that H2 is the limiting reactant. 2.13 g H2 should theoretically be able to produce:

2.13 g H2/2.00g/mol X 2 mol HCl/1 mol H2 X 36.45 g HCl/mol = 77.64 grams HCl.

Since the problem says you produced 74.30 g, your % yield is:

74.30/77.64 X 100 = 95.7%

Answer 2

moles H2 = 2.13 g/2 = 1.065

each mole H2 yields two moles HCl = 2.13 moles

HCl MW = 36.45

2.13 x 36.45 = 77.64 g

yield 74.30/77.64 x 100 = 95.70%

yes there is excess Cl, H2 is the limiting reactant with a percent yield of 95.70%