What weight of sodium chloride would be decomposed by `4.9 g` of sulphric acid, if `6 g` of sodium bisulphate `(NaHSO_(4))` and `1.825 g` of hydrogen chloride were produced in the reaction

Explanation

To solve the problem, we need to determine the weight of sodium chloride (NaCl) that would be decomposed by 4.9 g of sulfuric acid (H₂SO₄) given the products formed in the reaction.

### Step-by-Step Solution:

1. **Write the Balanced Chemical Equation:**
The reaction between sodium chloride and sulfuric acid can be represented as:
\[ \text{NaCl} + \text{H}_2\text{SO}_4 \rightarrow \text{NaHSO}_4 + \text{HCl} \]

2. **Identify the Given Data:**
- Weight of sulfuric acid (H₂SO₄) = 4.9 g
- Weight of sodium bisulfate (NaHSO₄) produced = 6 g
- Weight of hydrogen chloride (HCl) produced = 1.825 g

3. **Calculate the Total Weight of Products:**
The total weight of products formed in the reaction can be calculated as:
\[ \text{Weight of products} = \text{Weight of NaHSO}_4 + \text{Weight of HCl} = 6 \, \text{g} + 1.825 \, \text{g} = 7.825 \, \text{g} \]

4. **Apply the Law of Conservation of Mass:**
According to the law of conservation of mass, the total weight of reactants must equal the total weight of products. Therefore:
\[ \text{Weight of reactants} = \text{Weight of NaCl} + \text{Weight of H}_2\text{SO}_4 \]
Let the weight of NaCl be \( X \). Thus, we can write:
\[ X + 4.9 \, \text{g} = 7.825 \, \text{g} \]

5. **Solve for the Weight of NaCl (X):**
Rearranging the equation gives:
\[ X = 7.825 \, \text{g} - 4.9 \, \text{g} = 2.925 \, \text{g} \]

6. **Conclusion:**
The weight of sodium chloride that would be decomposed by 4.9 g of sulfuric acid is:
\[ \boxed{2.925 \, \text{g}} \]