The balance chemical equation is
2 #C_2##H_6# + 7#O_2# ---> 4C#O_2# + 6#H_2#O
as per the equation : 2 moles of #C_2##H_6# needs 7 moles of #O_2# .
moles of #C_2##H_6# = Volume of #C_2##H_6# / 22.4 L
moles of #C_2##H_6# = 16.4 L / 22.4 L = 0.73 mol
as per the molar ratio X mol of #C_2##H_6# will need react with 0.98 mol of #O_2#
2 mol of #C_2##H_6# /7 mol of #O_2# =
X mol of #C_2##H_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_2##H_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_2##H_6# hence it is a limiting reagent. 0.73 - 0.28 = 0.45 moles of #C_2##H_6# will remain unused , so it is an excess reagent.
Unused mass of #C_2##H_6# =
Unused moles of #C_2##H_6# x molar mass of #C_2##H_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#
