50 litres of water containing `Ca(HCO_3)_2` when converted into soft water required 22.2 g of `Ca(OH)_2`. Calculate the amount of `Ca(HCO_3)_2` per litre of hard water.

Explanation

To solve the problem, we need to determine the amount of calcium bicarbonate, \( \text{Ca(HCO}_3\text{)}_2 \), present in 1 liter of hard water, given that 50 liters of this water required 22.2 g of calcium hydroxide, \( \text{Ca(OH)}_2 \), to convert it into soft water.

### Step-by-step Solution:

1. **Write the Chemical Reaction:**
The reaction between calcium bicarbonate and calcium hydroxide can be represented as follows:
\[ \text{Ca(HCO}_3\text{)}_2 + \text{Ca(OH)}_2 \rightarrow 2 \text{CaCO}_3 + 2 \text{H}_2\text{O} \]
This shows that 1 mole of \( \text{Ca(HCO}_3\text{)}_2 \) reacts with 1 mole of \( \text{Ca(OH)}_2 \) to produce 2 moles of calcium carbonate and 2 moles of water.

2. **Calculate the Moles of \( \text{Ca(OH)}_2 \):**
The molar mass of \( \text{Ca(OH)}_2 \) is 74 g/mol. To find the number of moles of \( \text{Ca(OH)}_2 \) in 22.2 g:
\[ \text{Number of moles of } \text{Ca(OH)}_2 = \frac{\text{mass}}{\text{molar mass}} = \frac{22.2 \, \text{g}}{74 \, \text{g/mol}} \approx 0.3 \, \text{moles} \]

3. **Determine the Moles of \( \text{Ca(HCO}_3\text{)}_2 \):**
From the balanced equation, we see that 1 mole of \( \text{Ca(OH)}_2 \) reacts with 1 mole of \( \text{Ca(HCO}_3\text{)}_2 \). Therefore, 0.3 moles of \( \text{Ca(OH)}_2 \) will react with 0.3 moles of \( \text{Ca(HCO}_3\text{)}_2 \).

4. **Calculate the Mass of \( \text{Ca(HCO}_3\text{)}_2 \):**
The molar mass of \( \text{Ca(HCO}_3\text{)}_2 \) is 162 g/mol. The mass of 0.3 moles of \( \text{Ca(HCO}_3\text{)}_2 \) can be calculated as follows:
\[ \text{Mass of } \text{Ca(HCO}_3\text{)}_2 = \text{number of moles} \times \text{molar mass} = 0.3 \, \text{moles} \times 162 \, \text{g/mol} = 48.6 \, \text{g} \]

5. **Calculate the Amount of \( \text{Ca(HCO}_3\text{)}_2 \) per Liter of Hard Water:**
Since this mass (48.6 g) is present in 50 liters of water, we can find the concentration per liter:
\[ \text{Amount of } \text{Ca(HCO}_3\text{)}_2 \text{ per liter} = \frac{48.6 \, \text{g}}{50 \, \text{L}} = 0.972 \, \text{g/L} \]

### Final Answer:
The amount of \( \text{Ca(HCO}_3\text{)}_2 \) per liter of hard water is approximately **0.972 g/L**.