What is the percent yield when 7.80 grams of benzene burns in oxygen gas to form 3 grams of CO2 and water vapor?

C6H6 + 15 O2 = 12 CO2 + 6 H2O

moles of C6H6 = 7.8 / 78 = .1 moles

the O2 is in excess so benzene is the limiting agent and forms .1 x 6 moles of CO2

.6 moles of CO2 = .6 x 44 = 26.7 g at 100 % yield

but we have 3 g or 3 / 26.7 % = 11.23 %