The ratio of mass percentage (w/w) of C : H in a hydrocarbon is $12 : 1$. It has two carbon atoms. The weight (in g) of $CO_2(g)$ formed when $3.38$ g of this hydrocarbon is completely burnt in oxygen is : (Given : Molar mass in g mol$^{-1}$ C : 12, H : 1, O : 16)

Ratio of mass percentage of C : H = $12 : 1$

Ratio of moles of C : H = $\dfrac{12}{12} : \dfrac{1}{1} = 1 : 1$

The empirical formula is $CH$. Since the hydrocarbon has two carbon atoms, its molecular formula is $C_2H_2$.

Molar mass of $C_2H_2 = 2 \times 12 + 2 \times 1 = 26$ g mol$^{-1}$

Moles of $C_2H_2$ in $3.38$ g = $\dfrac{3.38}{26} = 0.13$ mol

The combustion reaction is:

$C_2H_2 + \dfrac{5}{2}O_2 \rightarrow 2CO_2 + H_2O$

From the stoichiometry, $1$ mole of $C_2H_2$ produces $2$ moles of $CO_2$.

Moles of $CO_2$ produced = $2 \times 0.13 = 0.26$ mol

Molar mass of $CO_2 = 12 + 2 \times 16 = 44$ g mol$^{-1}$

Weight of $CO_2$ formed = $0.26 \times 44 = 11.44$ g

Answer: $11.44$