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)

Let the empirical formula of the hydrocarbon be $$C_2H_n$$ because the problem states that it contains two carbon atoms.

Step 1:

Given mass-percentage ratio of carbon to hydrogen is $$12:1$$ (w/w).
For one mole of $$C_2H_n$$:
mass of carbon $$= 2 \times 12 = 24 \text{ g}$$
mass of hydrogen $$= n \times 1 = n \text{ g}$$

Using the given ratio,

$$\frac{24}{n} = \frac{12}{1} \quad \Longrightarrow \quad 24 = 12n \quad \Longrightarrow \quad n = 2.$$

Hence the molecular formula is $$C_2H_2$$ (ethyne/acetylene). 

Step 2:

The molar mass of $$C_2H_2$$ is
$$M_{C_2H_2}=2(12)+2(1)=26 \text{ g mol}^{-1}.$$

Moles of hydrocarbon present in 3.38 g sample:

$$n_{C_2H_2} = \frac{3.38}{26}=0.13 \text{ mol}.$$

Step 3:

Complete combustion reaction:

$$C_2H_2 + \frac{5}{2}O_2 \;\rightarrow\; 2CO_2 + H_2O.$$

From the stoichiometry, 

$$1$$ mole of $$C_2H_2$$ produces $$2$$ moles of $$CO_2$$.

Step 4:

Therefore, moles of $$CO_2$$ produced:

$$n_{CO_2}=0.13 \times 2 = 0.26 \text{ mol}.$$

Mass of $$CO_2$$ formed:

$$m_{CO_2}=n_{CO_2}\times M_{CO_2}=0.26 \times 44 = 11.44 \text{ g}.$$

Thus, the weight of carbon dioxide produced is $$11.44$$ g.

Option B $$\longrightarrow\ $$ $$11.44$$