If 2 CH3OH + 3 O2 -> 2CO2 + 4 H2O was carried out in the laboratory and 219 g of water was produced, what would the percent yield be if the reaction started with 229 g of methanol in excess oxygen gas

Question

If 2 CH3OH + 3 O2 -> 2CO2 + 4 H2O was carried out in the laboratory and 219 g of water was produced, what would the percent yield be if the reaction started with 229 g of methanol in excess oxygen gas

Answer 1

2CH₃OH + 3O₂ ==> 2CO₂ + 4H₂O

This balanced equation tells you that 2 moles of CH₃OH in excess O₂ can produce 4 moles H₂O.

So, we are given grams of H₂O produced (not moles), but we can convert g to moles. Let us do that.

219 g H₂O x 1 mol H₂O/18 g = 12.17 = 12.2 moles H₂O produced. This is the actual yield.

Now we need to know the theoretical yield...

229 g CH₃OH x 1 mol CH₃OH/32.0 g = 7.16 moles CH₃OH

7.16 moles CH₃OH x 4 moles H₂O/2 moles CH₃OH = 14.3 moles H₂O produced. This is the theoretical yield.

% yield is actual/theoretical (x100%) = 12.2 moles/14.3 moles (x100%) = 85.3% yield