The balance chemical equation is
2 C2H6 + 7O2 ---> 4CO2 + 6H2O
as per the equation : 2 moles of C2H6 needs 7 moles of O2 .
moles of C2H6 = Volume of C2H6 / 22.4 L
moles of C2H6 = 16.4 L / 22.4 L = 0.73 mol
as per the molar ratio X mol of C2H6 will need react with 0.98 mol of O2
2 mol of C2H6 /7 mol of O2 =
X mol of C2H6 / 0.98 mol of O2
7.x = 0.98 x 2
7x = 1.96 , x = 1.96/ 7 = 0.28 mol
0.28 moles of C2H6 can react with 0.98 mol of O2.
All of the oxygen will be used to react with 0.28 mol of C2H6 hence it is a limiting reagent. 0.73 - 0.28 = 0.45 moles of C2H6 will remain unused , so it is an excess reagent.
Unused mass of C2H6 =
Unused moles of C2H6 x molar mass of C2H6
= 0.45 mol x 30 g molโ1 = 13.5 g.
As for the volume of CO2 produced, we know from the balanced equation that 2 moles of C2H6 will produce 4 moles of CO2; thus, the moles of CO2 produced will be
nCO2=0.28โ
2=0.56 moles
Therefore, from n=VVmol, we have
V=22.4โ
0.56=12.54L