C6H6
C4H10
C5H10
C4H6
Correct answer is D
CxHy + (x + y/4)O2 → x CO2 + y/2 H2O
89.6 dm3 → x moles CO2
22.4 dm3 → 1 mole CO2
x = 89.622.4
= 4 moles CO2
54g of H2O → k moles H2O
18g H2O → 1 mole H2O
k = 5418
= 3 moles H2O
From the equation above,
x = 4
k = y/2 = 3
y/2 = 3 ⟹ y = 6
∴ The hydrocarbon C_xH_y = C_4H_6
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