How many mg of quick line are required to remove hardness from 1 kg of hard water having 366 ppm `HCO_(3)^(-) and Ca^(2+)` as the only cation:

Options

A. 72 mg
B. 84 mg
C. 168 mg
D. 170 mg

Answer

To solve the problem of how many mg of quick lime (calcium oxide, CaO) are required to remove hardness from 1 kg of hard water containing 366 ppm of bicarbonate ions (HCO₃⁻) and calcium ions (Ca²⁺), we can follow these steps:

### Step 1: Calculate the mass of bicarbonate ions in 1 kg of hard water.
Given that the concentration of bicarbonate ions is 366 ppm, we can use the formula for ppm:

\[ \text{ppm} = \frac{\text{mass of solute}}{\text{mass of solution}} \times 10^6 \]

Rearranging the formula to find the mass of the solute (bicarbonate ions):

\[ \text{mass of bicarbonate} = \text{ppm} \times \frac{\text{mass of solution}}{10^6} \]

Substituting the values:

\[ \text{mass of bicarbonate} = 366 \times \frac{1000 \text{ g}}{10^6} = 0.366 \text{ g} \]

### Step 2: Calculate the number of moles of bicarbonate ions.
To find the number of moles, we need the molar mass of bicarbonate ions (HCO₃⁻):

- Hydrogen (H) = 1 g/mol
- Carbon (C) = 12 g/mol
- Oxygen (O) = 16 g/mol × 3 = 48 g/mol

Thus, the molar mass of bicarbonate (HCO₃⁻) is:

\[ \text{Molar mass of HCO₃⁻} = 1 + 12 + 48 = 61 \text{ g/mol} \]

Now, we can calculate the number of moles of bicarbonate:

\[ \text{moles of bicarbonate} = \frac{\text{mass of bicarbonate}}{\text{molar mass of HCO₃⁻}} = \frac{0.366 \text{ g}}{61 \text{ g/mol}} \approx 0.006 \text{ moles} \approx 6 \text{ millimoles} \]

### Step 3: Determine the stoichiometry of the reaction.
The reaction between calcium ions (from quick lime) and bicarbonate ions is:

\[ \text{Ca}^{2+} + 2 \text{HCO}_3^{-} \rightarrow \text{Ca(HCO}_3)_2 \]

From the stoichiometry, we see that 1 mole of Ca²⁺ reacts with 2 moles of HCO₃⁻. Therefore, for 6 millimoles of HCO₃⁻:

\[ \text{moles of Ca}^{2+} = \frac{6 \text{ millimoles}}{2} = 3 \text{ millimoles} \]

### Step 4: Calculate the amount of quick lime (CaO) needed.
Since 1 mole of Ca²⁺ comes from 1 mole of CaO, we need 3 millimoles of CaO.

Now, we need the molar mass of CaO:

- Calcium (Ca) = 40 g/mol
- Oxygen (O) = 16 g/mol

Thus, the molar mass of CaO is:

\[ \text{Molar mass of CaO} = 40 + 16 = 56 \text{ g/mol} \]

### Step 5: Calculate the mass of quick lime required.
Using the number of moles and the molar mass:

\[ \text{mass of CaO} = \text{moles of CaO} \times \text{molar mass of CaO} = 3 \text{ millimoles} \times 56 \text{ g/mol} \]

Converting millimoles to grams:

\[ \text{mass of CaO} = 3 \times 0.056 \text{ g} = 0.168 \text{ g} = 168 \text{ mg} \]

### Final Answer
Therefore, the amount of quick lime required to remove hardness from 1 kg of hard water is **168 mg**.

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