Complete combustion of 3 g of ethane gives x $$\times 10^{22}$$ molecules of water. The value of x is ________ (Round off to the Nearest Integer).
[Use: $$N_A = 6.023 \times 10^{23}$$; Atomic masses in u: C: 12.0; O: 16.0; H: 1.0]

The balanced equation for the complete combustion of ethane is: $$2\text{C}_2\text{H}_6 + 7\text{O}_2 \rightarrow 4\text{CO}_2 + 6\text{H}_2\text{O}$$. From this equation, 2 moles of ethane produce 6 moles of water, so 1 mole of ethane produces 3 moles of water.

The molar mass of ethane ($$\text{C}_2\text{H}_6$$) is $$2(12) + 6(1) = 30$$ g/mol. The number of moles of ethane in 3 g is: $$n = \frac{3}{30} = 0.1 \text{ mol}$$.

Since each mole of ethane produces 3 moles of water, the moles of water produced are: $$0.1 \times 3 = 0.3 \text{ mol}$$.

The number of water molecules is: $$N = 0.3 \times 6.023 \times 10^{23} = 1.8069 \times 10^{23}$$. This can be written as $$18.069 \times 10^{22}$$.

Since $$N = x \times 10^{22}$$, we get $$x = 18.069$$. Rounding to the nearest integer, $$x = 18$$.