CxHy + (x + y/4)O2 to x CO2 + y/2 H2O
89.6 dm3 to x moles CO2
22.4 dm3 to 1 mole CO2
x = 89.622.4
= 4 moles CO2
54g of H2O to k moles H2O
18g H2O to 1 mole H2O
k = 5418
= 3 moles H2O
From the equation above,
x = 4
k = y/2 = 3
y/2 = 3 implies y = 6
∴ The hydrocarbon CxHy = C4H6
