JAMB Chemistry 2019 Paper

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 = 4

k = y/2 = 3

y/2 = 3 $\implies$ y = 6

$\therefore$ The hydrocarbon C$_{x}$H$_{y}$ = C$_{4}$H$_{6}$