C\(_x\)H\(_y\) + (x + y/4)O\(_2\) \(\to\) x CO\(_2\) + y/2 H\(_2\)O
89.6 dm\(^3\) \(\to\) x moles CO\(_2\)
22.4 dm\(^3\) \(\to\) 1 mole CO\(_2\)
x = \(\frac{89.6}{22.4}\)
= 4 moles CO\(_2\)
54g of H\(_2\)O \(\to\) k moles H\(_2\)O
18g H\(_2\)O \(\to\) 1 mole H\(_2\)O
k = \(\frac{54}{18}\)
= 3 moles H\(_2\)O
From the equation above,
x = 4Â
k = y/2 = 3
y/2 = 3 \(\implies\) y = 6
\(\therefore\) The hydrocarbon C\(_x\)H\(_y\) = C\(_4\)H\(_6\)