1. Write the combustion reaction of ethane

$\mathrm{C_2H_6 + \frac{7}{2}O_2 \rightarrow 2CO_2 + 3H_2O}$

From the balanced equation:

• $1$ mole of ethane gives $3$ moles of water.

2. Find moles of ethane in 3 g

Molar mass of ethane, $\mathrm{C_2H_6}$:

$2(12) + 6(1) = 24 + 6 = 30\ \text{g mol}^{-1}$

So, moles of ethane in $3$ g:

$n(\mathrm{C_2H_6}) = \frac{3}{30} = 0.1\ \text{mol}$

3. Find moles of water formed

Since $1$ mole ethane gives $3$ moles water,

$n(\mathrm{H_2O}) = 0.1 \times 3 = 0.3\ \text{mol}$

4. Convert moles of water to number of molecules

Using Avogadro number,

$N = 0.3 \times 6.023 \times 10^{23}$

$N = 1.8069 \times 10^{23}$

Now write this as $x \times 10^{22}$:

$1.8069 \times 10^{23} = 18.069 \times 10^{22}$

So,

$x = 18.069$

Rounded to the nearest integer:

$x = 18$

5. Final Answer

$\boxed{18}$

6. Comparison with stored correct answer

Stored correct answer = $18$.

My derived answer also is $18$, so they agree.