How many grams of CO2 could be produced from the combustion of 0.0713 mol. of C2H4 in the presence of 0.156 mol. of O2?

Answer 1

C2H4 + 3O2 --> 2CO2 + 2H2O ideally
so for .0713 mol C2H4 we need 3*.0713 = .213 mol O2
so we do not have enough O2
so .156 mol O2 is limiting
for .156 mol O2 we get
(2/3).156 mol of CO2
or
.104 mol of CO2
C = 12 g/mol
O2 = 32 g/mol
so
CO2 = 44 g/mol

44 * .104 = 4.58 grams of CO2