$0.144\text{ g}$ of an aluminium compound X is reacted with excess water, producing a gas. This gas then burns completely in $\text{O}_2$ to give $\text{H}_2\text{O}$ and only $72\text{ cm}^3$ of $\text{CO}_2$. The volume of $\text{CO}_2$ was recorded at room temperature and pressure. What could be the formula of X? [$\text{C} = 12.0$, $\text{Al} = 27.0$; $1$ mole of any gas occupies $24\text{ dm}^3$ at room temperature and pressure]
- A$\text{Al}_2\text{C}_3$
- B$\text{Al}_3\text{C}_4$
- C$\text{Al}_4\text{C}_3$
- D$\text{Al}_5\text{C}_3$