Asked by Jass
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?
Answers
Answered by
Damon
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
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
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