What is the percent of yield

Question

Ethane is combustible. suppose 4.0 moles of ethane and 5.0 moles of oxygen are mixed for combustion. What is the % yield of water if 43 grams of water are produced and collected from this combustion reaction?

Answer 1

The balanced equation is:

2(C2H6) + 7(O2) --> 4(CO2) + 6(H2O)

We determine that 12 mol H2O can be made from 4 mol C2H6:

4 mol C2H6 * 6 mol H2O/2 mol C2H6 = 12 mol H20

however only 4.28 mole H2O can be made from 5 mol of O2:

5 mol O2 * 6 mol H2)/7 mol O2 = 4.28 mol H2O

therefore oxygen is the limiting reactant and only 4.28 mol of O2 can be made in the combustion. then convert to grams

4.28 mol H2O * 18 g H2O/1 mol H2O = 77.04 g H2O which is the theoretical yield

the problem says 43g of H2O is produced which is the actual yield

percent yield = actual yield/theoretical yield = 43g/77.04g = 55.8%