41.6 g `BaCl_(2)` react with how many grams of sodium sulphate to produce 46.6 g of barium sulphate and 23.4 g of sodium chloride ?

Options

A. 48.8 g
B. 28.4 g
C. 12.2 g
D. 45.6 g

Answer

To solve the problem, we need to determine how many grams of sodium sulfate (Na₂SO₄) are required to react with 41.6 g of barium chloride (BaCl₂) to produce 46.6 g of barium sulfate (BaSO₄) and 23.4 g of sodium chloride (NaCl).

### Step-by-Step Solution:

1. **Write the Balanced Chemical Equation**:
The reaction between barium chloride and sodium sulfate can be represented as follows:
\[ \text{BaCl}_2 + \text{Na}_2\text{SO}_4 \rightarrow \text{BaSO}_4 + 2 \text{NaCl} \]
This equation shows that 1 mole of BaCl₂ reacts with 1 mole of Na₂SO₄ to produce 1 mole of BaSO₄ and 2 moles of NaCl.

2. **Calculate the Total Mass of Products**:
We can find the total mass of the products formed:
\[ \text{Total mass of products} = \text{mass of BaSO}_4 + \text{mass of NaCl} \]
\[ = 46.6 \, \text{g} + 23.4 \, \text{g} = 70.0 \, \text{g} \]

3. **Apply the Law of Conservation of Mass**:
According to the law of conservation of mass, the total mass of the reactants must equal the total mass of the products. Therefore:
\[ \text{Total mass of reactants} = \text{mass of BaCl}_2 + \text{mass of Na}_2\text{SO}_4 \]
Let the mass of Na₂SO₄ be \( x \) grams. Thus, we have:
\[ 41.6 \, \text{g} + x = 70.0 \, \text{g} \]

4. **Solve for \( x \)**:
Rearranging the equation gives:
\[ x = 70.0 \, \text{g} - 41.6 \, \text{g} \]
\[ x = 28.4 \, \text{g} \]

5. **Conclusion**:
Therefore, the mass of sodium sulfate required is \( 28.4 \, \text{g} \).

### Final Answer:
The mass of sodium sulfate (Na₂SO₄) required is **28.4 g**.

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