A solution of ammonia in water with a density of 0.93 g/cm3 contained 18% by mass of ammonia. What volume of this solution is required to form 30g of ammonium tetraoxosulphate(VI) based on the reactio

Molar mass of $(\text{NH}_4)_2\text{SO}_4 = 2(\text{N}) + 8(\text{H}) + 1(\text{S}) + 4(\text{O})$

$$= 2(14.01 \text{ g/mol}) + 8(1.01 \text{ g/mol}) + 32.07 \text{ g/mol} + 4(16.00 \text{ g/mol})$$ $$= 28.02 + 8.08 + 32.07 + 64.00 = 132.17 \text{ g/mol}$$

$$\text{Moles} = \frac{\text{Mass}}{\text{Molar mass}}$$ $$\text{Moles of } (\text{NH}_4)_2\text{SO}_4 = \frac{30 \text{ g}}{132.17 \text{ g/mol}} = 0.22698 \text{ mol}$$

The balanced equation is: $2\text{NH}_3 + \text{H}_2\text{SO}_4 \rightarrow (\text{NH}_4)_2\text{SO}_4$.
From the equation, 2 moles of NH$_3$ react to form 1 mole of $(\text{NH}_4)_2\text{SO}_4$.

$$\text{Moles of NH}_3 = 0.22698 \text{ mol } (\text{NH}_4)_2\text{SO}_4 \times \frac{2 \text{ mol NH}_3}{1 \text{ mol } (\text{NH}_4)_2\text{SO}_4}$$ $$= 0.45396 \text{ mol NH}_3$$

Molar mass of NH$_3 = 1(\text{N}) + 3(\text{H})$

$$= 14.01 \text{ g/mol} + 3(1.01 \text{ g/mol}) = 14.01 + 3.03 = 17.04 \text{ g/mol}$$ $$\text{Mass of NH}_3 = \text{Moles} \times \text{Molar mass}$$ $$= 0.45396 \text{ mol} \times 17.04 \text{ g/mol} = 7.735 \text{ g}$$

The solution contains 18% by mass of ammonia.

$$\text{Mass of solution} = \frac{\text{Mass of pure NH}_3}{\text{Fractional abundance of NH}_3}$$ $$= \frac{7.735 \text{ g}}{0.18} = 42.972 \text{ g}$$

The density of the ammonia solution is $0.93 \text{ g/cm}^3$, which is equivalent to $0.93 \text{ g/mL}$.

$$\text{Volume of solution} = \frac{\text{Mass of solution}}{\text{Density of solution}}$$ $$= \frac{42.972 \text{ g}}{0.93 \text{ g/mL}} = 46.206 \text{ mL}$$

Rounding to three significant figures:

$$\text{Volume of solution} = 46.2 \text{ mL}$$

The volume of the ammonia solution required is $\text{46.2 mL}$.