Sure, let's go through the steps to determine the volume of the ammonia solution required to produce 30 g of ammonium tetraoxosulphate(VI) [tex]\(\left( NH _4\right)_2 SO _4\)[/tex].
### 1. Moles of [tex]\(\left( NH _4\right)_2 SO _4\)[/tex]
First, we need to find the moles of [tex]\(\left( NH _4\right)_2 SO _4\)[/tex] produced given the mass.
Given:
- Final mass of [tex]\(\left( NH _4\right)_2 SO _4\)[/tex] = 30 g
- Molar mass of [tex]\(\left( NH _4\right)_2 SO _4\)[/tex] = 132.14 g/mol
The moles of [tex]\(\left( NH _4\right)_2 SO _4\)[/tex] produced:
[tex]\[ \text{Moles of } \left(NH_4\right)_2SO_4 = \frac{\text{Mass of }\left(NH_4\right)_2SO_4}{\text{Molar mass of }\left(NH_4\right)_2SO_4} = \frac{30 \text{ g}}{132.14 \text{ g/mol}} \approx 0.227 \text{ moles} \][/tex]
### 2. Moles of NH[tex]\(_3\)[/tex] Required
The reaction states that 2 moles of NH[tex]\(_3\)[/tex] are required to produce 1 mole of [tex]\(\left( NH _4\right)_2 SO _4\)[/tex].
So, the moles of NH[tex]\(_3\)[/tex] required:
[tex]\[ \text{Moles of } NH_3 = 2 \times \text{Moles of } \left(NH_4\right)_2SO_4 = 2 \times 0.227 \approx 0.454 \text{ moles} \][/tex]
### 3. Mass of NH[tex]\(_3\)[/tex] Required
Next, we need to find the mass of NH[tex]\(_3\)[/tex] required.
Given:
- Molar mass of NH[tex]\(_3\)[/tex] = 17.03 g/mol
[tex]\[ \text{Mass of } NH_3 = \text{Moles of } NH_3 \times \text{Molar mass of } NH_3 = 0.454 \times 17.03 \approx 7.733 \text{ g} \][/tex]
### 4. Mass of the Solution Required
We know that the solution contains 18% ammonia by mass.
[tex]\[ \text{Mass of ammonia solution required} = \frac{\text{Mass of } NH_3}{\text{Mass fraction of } NH_3} = \frac{7.733 \text{ g}}{0.18} \approx 42.959 \text{ g} \][/tex]
### 5. Volume of the Solution Required
Finally, we use the density of the solution to find the volume of the solution required.
Given:
- Density of the solution = 0.93 g/cm[tex]\(^3\)[/tex]
[tex]\[ \text{Volume of solution required} = \frac{\text{Mass of solution required}}{\text{Density of solution}} = \frac{42.959 \text{ g}}{0.93 \text{ g/cm}^3} \approx 46.193 \text{ cm}^3 \][/tex]
### Conclusion
The volume of the ammonia solution required to produce 30 g of ammonium tetraoxosulphate(VI) is approximately 46.193 cm[tex]\(^3\)[/tex].
