The balance chemical equation is
2 C_2H_6 + 7O_2 ---> 4CO_2 + 6H_2O
as per the equation : 2 moles of C_2H_6 needs 7 moles of O_2 .
moles of C_2H_6 = Volume of C_2H_6 / 22.4 L
moles of C_2H_6 = 16.4 L / 22.4 L = 0.73 mol
as per the molar ratio X mol of C_2H_6 will need react with 0.98 mol of O_2
2 mol of C_2H_6 /7 mol of O_2 =
X mol of C_2H_6 / 0.98 mol of O_2
7.x = 0.98 x 2
7x = 1.96 , x = 1.96/ 7 = 0.28 mol
0.28 moles of C_2H_6 can react with 0.98 mol of O_2.
All of the oxygen will be used to react with 0.28 mol of C_2H_6 hence it is a limiting reagent. 0.73 - 0.28 = 0.45 moles of C_2H_6 will remain unused , so it is an excess reagent.
Unused mass of C_2H_6 =
Unused moles of C_2H_6 x molar mass of C_2H_6
= 0.45 mol x 30 g mol^(-1) = 13.5 g.
As for the volume of CO_2 produced, we know from the balanced equation that 2 moles of C_2H_6 will produce 4 moles of CO_2; thus, the moles of CO_2 produced will be
n_(CO_2) = 0.28 * 2 = 0.56 moles
Therefore, from n = V/V_(mol), we have
V = 22.4 * 0.56 = 12.54 L
