What mass of sodium chloride would be decomposed by 9.8 g of sulphuric acid if 12 g of sodium bisulphate and 2.75 g of hydrogen chloride were produced in a reaction?

Options

A. 14.75 g
B. 3.8 g
C. 4.95 g
D. 2.2 g

Explanation

To solve the problem, we need to analyze the reaction between sodium chloride (NaCl) and sulfuric acid (H2SO4) and determine how much sodium chloride is decomposed based on the products formed. The reaction can be represented as follows:

\[ \text{2 NaCl} + \text{H2SO4} \rightarrow \text{Na2SO4} + \text{2 HCl} \]

### Step 1: Identify the Molar Masses
First, we need to calculate the molar masses of the substances involved:
- Molar mass of NaCl = 22.99 (Na) + 35.45 (Cl) = 58.44 g/mol
- Molar mass of H2SO4 = 2(1.01) (H) + 32.07 (S) + 4(16.00) (O) = 98.09 g/mol
- Molar mass of Na2SO4 = 2(22.99) + 32.07 + 4(16.00) = 142.05 g/mol
- Molar mass of HCl = 1.01 (H) + 35.45 (Cl) = 36.46 g/mol

### Step 2: Calculate Moles of Products
Next, we calculate the moles of the products formed:
- Moles of sodium bisulfate (Na2SO4) produced:
\[ \text{Moles of Na2SO4} = \frac{\text{mass}}{\text{molar mass}} = \frac{12 \, \text{g}}{142.05 \, \text{g/mol}} \approx 0.0845 \, \text{mol} \]

- Moles of hydrogen chloride (HCl) produced:
\[ \text{Moles of HCl} = \frac{2.75 \, \text{g}}{36.46 \, \text{g/mol}} \approx 0.0755 \, \text{mol} \]

### Step 3: Determine the Stoichiometry
From the balanced equation, we see that:
- 2 moles of NaCl produce 2 moles of HCl.
- Therefore, the moles of NaCl required to produce 0.0755 moles of HCl is also 0.0755 moles.

### Step 4: Calculate Mass of Sodium Chloride
Now, we can calculate the mass of sodium chloride that would be decomposed:
\[ \text{Mass of NaCl} = \text{moles} \times \text{molar mass} = 0.0755 \, \text{mol} \times 58.44 \, \text{g/mol} \approx 4.41 \, \text{g} \]

### Step 5: Final Calculation
Since the question asks for the mass of sodium chloride decomposed, we can conclude:
\[ \text{Mass of NaCl decomposed} \approx 4.41 \, \text{g} \]

### Conclusion
Thus, the mass of sodium chloride that would be decomposed by 9.8 g of sulfuric acid is approximately **4.41 g**.

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