Calculate the mass of `Na_(2)CO_(3)` which will have the same number of molecules as contained in 12.3 g of `MgSO_(4).7 H_(2)O`.

Explanation

To solve the problem of calculating the mass of Na₂CO₃ that has the same number of molecules as contained in 12.3 g of MgSO₄·7H₂O, we can follow these steps:

### Step 1: Calculate the molar mass of MgSO₄·7H₂O
- The molar mass of MgSO₄·7H₂O can be calculated as follows:
- Molar mass of Mg = 24.31 g/mol
- Molar mass of S = 32.07 g/mol
- Molar mass of O = 16.00 g/mol (4 O in SO₄)
- Molar mass of H₂O = 18.02 g/mol (7 H₂O)

\[ \text{Molar mass of MgSO₄} = 24.31 + 32.07 + (4 \times 16.00) = 120.37 \, \text{g/mol} \]

\[ \text{Molar mass of 7H₂O} = 7 \times 18.02 = 126.14 \, \text{g/mol} \]

\[ \text{Total molar mass of MgSO₄·7H₂O} = 120.37 + 126.14 = 246.51 \, \text{g/mol} \]

### Step 2: Calculate the number of moles of MgSO₄·7H₂O in 12.3 g
- Using the formula for moles:

\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{12.3 \, \text{g}}{246.51 \, \text{g/mol}} \approx 0.0499 \, \text{moles} \]

### Step 3: Find the number of moles of Na₂CO₃ needed
- Since we want the same number of molecules, we need the same number of moles of Na₂CO₃:

\[ \text{Moles of Na₂CO₃} = 0.0499 \, \text{moles} \]

### Step 4: Calculate the molar mass of Na₂CO₃
- The molar mass of Na₂CO₃ can be calculated as follows:
- Molar mass of Na = 22.99 g/mol (2 Na)
- Molar mass of C = 12.01 g/mol
- Molar mass of O = 16.00 g/mol (3 O)

\[ \text{Molar mass of Na₂CO₃} = (2 \times 22.99) + 12.01 + (3 \times 16.00) = 45.98 + 12.01 + 48.00 = 105.99 \, \text{g/mol} \]

### Step 5: Calculate the mass of Na₂CO₃ needed
- Using the formula for mass:

\[ \text{Mass of Na₂CO₃} = \text{moles} \times \text{molar mass} = 0.0499 \, \text{moles} \times 105.99 \, \text{g/mol} \approx 5.28 \, \text{g} \]

### Final Answer
The mass of Na₂CO₃ that will have the same number of molecules as contained in 12.3 g of MgSO₄·7H₂O is approximately **5.28 g**.

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