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) 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 can be represented as:
\[ \text{BaCl}_2 + \text{Na}_2\text{SO}_4 \rightarrow \text{BaSO}_4 + 2\text{NaCl} \]

2. **Identify the masses of the products:**
From the problem, we know:
- Mass of BaSO4 = 46.6 g
- Mass of NaCl = 23.4 g

3. **Calculate the total mass of the products:**
\[ \text{Total mass of products} = \text{Mass of BaSO}_4 + \text{Mass of NaCl} = 46.6 \, \text{g} + 23.4 \, \text{g} = 70.0 \, \text{g} \]

4. **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:
\[ \text{Mass of reactants} = \text{Mass of BaCl}_2 + \text{Mass of Na}_2\text{SO}_4 \]
Let \( X \) be the mass of sodium sulfate (Na2SO4). We have:
\[ 41.6 \, \text{g} + X = 70.0 \, \text{g} \]

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

6. **Conclusion:**
Therefore, the mass of sodium sulfate that reacts is **28.4 g**.