What quanitity of ammonium sulphate is necessary for the production of `NH_(3)` gas sufficient to neutralize a solution containing 292 g of HC1? `[HCl=36.5,(NH_(4))_(2) SO_(4)=132, NH_(3)=17]`

Options

A. 272 g
B. 403 g
C. 528 g
D. 1056 g

Explanation

To determine the quantity of ammonium sulfate necessary to produce enough ammonia gas to neutralize a solution containing 292 g of HCl, we can follow these steps:

### Step 1: Calculate the number of moles of HCl
To find the number of moles of HCl, we use the formula:

\[ \text{Number of moles} = \frac{\text{Given mass}}{\text{Molar mass}} \]

Given:
- Mass of HCl = 292 g
- Molar mass of HCl = 36.5 g/mol

\[ \text{Number of moles of HCl} = \frac{292 \, \text{g}}{36.5 \, \text{g/mol}} \approx 8 \, \text{moles} \]

### Step 2: Determine the number of moles of NH3 required
From the neutralization reaction, we know that:

\[ \text{HCl} + \text{NH}_3 \rightarrow \text{NH}_4\text{Cl} \]

This indicates that 1 mole of HCl reacts with 1 mole of NH3. Therefore, the number of moles of NH3 required is equal to the number of moles of HCl:

\[ \text{Number of moles of NH}_3 = 8 \, \text{moles} \]

### Step 3: Relate moles of NH3 to moles of (NH4)2SO4
From the reaction of ammonium sulfate with sodium hydroxide, we have:

\[ \text{(NH}_4)_2\text{SO}_4 + 2 \text{NaOH} \rightarrow 2 \text{NH}_3 + \text{Na}_2\text{SO}_4 + 2 \text{H}_2\text{O} \]

This shows that 1 mole of ammonium sulfate produces 2 moles of NH3. Therefore, to find the number of moles of ammonium sulfate needed to produce 8 moles of NH3, we can set up the following proportion:

\[ \text{Moles of (NH}_4)_2\text{SO}_4 = \frac{8 \, \text{moles of NH}_3}{2} = 4 \, \text{moles} \]

### Step 4: Calculate the mass of ammonium sulfate required
Now, we can calculate the mass of ammonium sulfate needed using its molar mass:

\[ \text{Mass} = \text{Number of moles} \times \text{Molar mass} \]

Given:
- Molar mass of (NH4)2SO4 = 132 g/mol

\[ \text{Mass of (NH}_4)_2\text{SO}_4 = 4 \, \text{moles} \times 132 \, \text{g/mol} = 528 \, \text{g} \]

### Conclusion
The quantity of ammonium sulfate necessary for the production of ammonia gas sufficient to neutralize a solution containing 292 g of HCl is **528 g**.

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