What weight of NaCI would be decomposed by 4.900 g of `H_2SO_4`, if 6 g of `NaHSO_4` and 1.825 g of HCI are produced in the reaction ?

Answer

To solve the problem, we will follow these steps:

### Step 1: Write the balanced chemical equation

The balanced chemical equation for the reaction between sodium chloride (NaCl) and sulfuric acid (H₂SO₄) is:

\[ \text{NaCl} + \text{H}_2\text{SO}_4 \rightarrow \text{NaHSO}_4 + \text{HCl} \]

### Step 2: Identify the given data

We are given:
- Mass of H₂SO₄ = 4.900 g
- Mass of NaHSO₄ produced = 6.000 g
- Mass of HCl produced = 1.825 g

### Step 3: Calculate the number of moles of the reactants and products

We need the molar masses of the compounds:
- Molar mass of H₂SO₄ = 98.08 g/mol
- Molar mass of NaHSO₄ = 120.06 g/mol
- Molar mass of HCl = 36.46 g/mol

Now, we can calculate the number of moles for each substance:

1. **Moles of H₂SO₄:**
\[ \text{Moles of H}_2\text{SO}_4 = \frac{\text{mass}}{\text{molar mass}} = \frac{4.900 \, \text{g}}{98.08 \, \text{g/mol}} \approx 0.0500 \, \text{mol} \]

2. **Moles of NaHSO₄:**
\[ \text{Moles of NaHSO}_4 = \frac{6.000 \, \text{g}}{120.06 \, \text{g/mol}} \approx 0.0500 \, \text{mol} \]

3. **Moles of HCl:**
\[ \text{Moles of HCl} = \frac{1.825 \, \text{g}}{36.46 \, \text{g/mol}} \approx 0.0500 \, \text{mol} \]

### Step 4: Relate the moles of reactants to the moles of products

From the balanced equation, we can see that:
- 1 mole of NaCl reacts with 1 mole of H₂SO₄ to produce 1 mole of NaHSO₄ and 1 mole of HCl.

Since we have 0.0500 moles of H₂SO₄, it will react with an equal amount of NaCl (0.0500 moles).

### Step 5: Calculate the mass of NaCl decomposed

Now we can calculate the mass of NaCl that would be decomposed:

1. **Molar mass of NaCl = 58.44 g/mol**
2. **Mass of NaCl = moles × molar mass**
\[ \text{Mass of NaCl} = 0.0500 \, \text{mol} \times 58.44 \, \text{g/mol} \approx 2.922 \, \text{g} \]

### Final Answer

The weight of NaCl that would be decomposed by 4.900 g of H₂SO₄ is approximately **2.92 g**.

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