Permanent hardness is due to `CI^(ɵ)` and `SO_4^(2-)` of `Mg^(2+)` and `Ca^(2+)` and is removed by adding `Na_2CO_3`.
`{:(CaSO_(4)+Na_(2)CO_(3)toCaCO_(3)+Na_(2)SO_(4)),(CaCl_(2)+Na_(2)CO_(3)toCaCO_(3)+2NaCl):}` Which of the following statements is`//`are correct?

Options

A. If hardness is 100 ppm `CaCO_(3)` the amount of `Na_(2)CO_(3)` requried to soften 10 L of hard water is 1.06 g.
B. If hardness is 100 ppm `CaCO_(3)`, the amount of `Na_(2)CO_(3)` required to soften 10 L of hard is 10.6 g
C. If hardness is 420 ppm `MgCO_(3)`, the amount of `Na_(2)CO_(3)` required to soften 10 L of hard water is 53.0 g.
D. If hardness is 420 ppm `MgCO_(3)` the amount of `Na_(2)CO_(3)` required to soften 10 L of hard water is 5.3 g.

Explanation

To solve the problem regarding the removal of permanent hardness in water caused by `Ca^(2+)` and `Mg^(2+)` ions, we need to analyze the provided chemical reactions and the given statements about the required amount of `Na2CO3` to remove the hardness.

### Step-by-Step Solution:

1. **Understanding the Problem:**
Permanent hardness in water is primarily due to the presence of `Ca^(2+)` and `Mg^(2+)` ions, which can be in the form of `CaSO4` and `MgSO4`. To remove this hardness, `Na2CO3` (sodium carbonate) is added.

2. **Chemical Reactions:**
The reactions provided in the question are:
- \( \text{CaSO}_4 + \text{Na}_2\text{CO}_3 \rightarrow \text{CaCO}_3 + \text{Na}_2\text{SO}_4 \)
- \( \text{CaCl}_2 + \text{Na}_2\text{CO}_3 \rightarrow \text{CaCO}_3 + 2\text{NaCl} \)

These reactions show that for every mole of `CaSO4` or `CaCl2`, one mole of `Na2CO3` is required to precipitate `CaCO3`, which is insoluble.

3. **Calculating the Amount of `Na2CO3` Needed:**
- Given that the hardness is represented as `100 ppm` of `CaCO3`, this means there are 100 grams of `CaCO3` in 1,000,000 mL (1,000 L) of water.
- For 10 liters of water, the amount of `CaCO3` can be calculated as:
\[ \text{Mass of } CaCO3 = \frac{100 \, \text{g}}{1,000,000 \, \text{mL}} \times 10,000 \, \text{mL} = 1 \, \text{g} \]

4. **Finding Moles of `CaCO3`:**
- The molar mass of `CaCO3` is approximately 100 g/mol.
- Therefore, the number of moles of `CaCO3` in 1 g is:
\[ \text{Moles of } CaCO3 = \frac{1 \, \text{g}}{100 \, \text{g/mol}} = 0.01 \, \text{mol} \]

5. **Moles of `Na2CO3` Required:**
- Since the stoichiometry of the reaction shows that 1 mole of `Na2CO3` is required for every mole of `CaCO3`, the moles of `Na2CO3` required will also be 0.01 mol.
- The molar mass of `Na2CO3` is approximately 106 g/mol.
- The mass of `Na2CO3` required is:
\[ \text{Mass of } Na2CO3 = 0.01 \, \text{mol} \times 106 \, \text{g/mol} = 1.06 \, \text{g} \]

6. **Calculating for `MgCO3`:**
- If the water contains `420 ppm` of `MgCO3`, this means there are 420 grams of `MgCO3` in 1,000,000 mL of water.
- For 10 liters, the amount of `MgCO3` can be calculated as:
\[ \text{Mass of } MgCO3 = \frac{420 \, \text{g}}{1,000,000 \, \text{mL}} \times 10,000 \, \text{mL} = 4.2 \, \text{g} \]

7. **Finding Moles of `MgCO3`:**
- The molar mass of `MgCO3` is approximately 84 g/mol.
- Therefore, the number of moles of `MgCO3` in 4.2 g is:
\[ \text{Moles of } MgCO3 = \frac{4.2 \, \text{g}}{84 \, \text{g/mol}} = 0.05 \, \text{mol} \]

8. **Moles of `Na2CO3` Required for `MgCO3`:**
- The moles of `Na2CO3` required will also be 0.05 mol.
- The mass of `Na2CO3` required is:
\[ \text{Mass of } Na2CO3 = 0.05 \, \text{mol} \times 106 \, \text{g/mol} = 5.3 \, \text{g} \]

9. **Conclusion:**
- From the calculations, we find that:
- For `CaCO3`, 1.06 g of `Na2CO3` is required.
- For `MgCO3`, 5.3 g of `Na2CO3` is required.
- Therefore, the correct statements regarding the amounts of `Na2CO3` needed are:
- 1.06 g for `CaCO3` (correct)
- 5.3 g for `MgCO3` (correct)

### Summary of Correct Statements:
- The correct amounts of `Na2CO3` required are:
- 1.06 grams for `CaCO3`
- 5.3 grams for `MgCO3`