10 g of a mixture of BaO and CaO requires 100 `cm^(3)` of 2.5M HCl to react completely. The percentage of calcium oxide in the mixture is approximately (Given molar mass of BaO = 153, CaO = 56)

Options

A. 52.6
B. 55.1
C. 44.9
D. 47.4

Explanation

To solve the problem, we need to determine the percentage of calcium oxide (CaO) in a mixture of barium oxide (BaO) and calcium oxide (CaO) given that 10 g of the mixture reacts with 100 cm³ of 2.5 M HCl.

### Step-by-Step Solution:

1. **Calculate the moles of HCl used:**
\[ \text{Molarity (M)} = \frac{\text{moles of solute}}{\text{volume of solution in liters}} \]
Given:
- Molarity of HCl = 2.5 M
- Volume of HCl = 100 cm³ = 0.1 L (since 1000 cm³ = 1 L)

\[ \text{Moles of HCl} = \text{Molarity} \times \text{Volume} = 2.5 \, \text{mol/L} \times 0.1 \, \text{L} = 0.25 \, \text{moles} \]

2. **Determine the stoichiometry of the reactions:**
The reactions are:
- For BaO:
\[ \text{BaO} + 2 \text{HCl} \rightarrow \text{BaCl}_2 + \text{H}_2\text{O} \]
- For CaO:
\[ \text{CaO} + 2 \text{HCl} \rightarrow \text{CaCl}_2 + \text{H}_2\text{O} \]

From the reactions, we see that:
- 1 mole of BaO reacts with 2 moles of HCl.
- 1 mole of CaO reacts with 2 moles of HCl.

Therefore, for a mixture of BaO and CaO, the total moles of HCl required for the mixture can be represented as:
\[ 2 \, \text{moles of HCl} \text{ for } 1 \, \text{mole of BaO} + 2 \, \text{moles of HCl} \text{ for } 1 \, \text{mole of CaO} \]

3. **Relate moles of HCl to moles of the mixture:**
Since 0.25 moles of HCl are used, we can set up the equation:
\[ 0.25 \, \text{moles of HCl} = 2 \times \text{moles of mixture} \]
Thus, the moles of the mixture (BaO + CaO) can be calculated as:
\[ \text{Moles of mixture} = \frac{0.25}{2} = 0.125 \, \text{moles} \]

4. **Set up the mass equation:**
Let \( x \) be the mass of CaO in grams. Then, the mass of BaO will be \( 10 - x \) grams.

The moles of CaO and BaO can be expressed as:
\[ \text{Moles of CaO} = \frac{x}{56} \]
\[ \text{Moles of BaO} = \frac{10 - x}{153} \]

The total moles of the mixture is:
\[ \frac{x}{56} + \frac{10 - x}{153} = 0.125 \]

5. **Solve for \( x \):**
Multiply through by the least common multiple (LCM) of 56 and 153 to eliminate the denominators:
\[ 153x + 56(10 - x) = 0.125 \times 56 \times 153 \]
Simplifying gives:
\[ 153x + 560 - 56x = 1071 \]
\[ 97x = 1071 - 560 \]
\[ 97x = 511 \]
\[ x = \frac{511}{97} \approx 5.27 \, \text{g} \]

6. **Calculate the percentage of CaO in the mixture:**
\[ \text{Percentage of CaO} = \left( \frac{x}{10} \right) \times 100 = \left( \frac{5.27}{10} \right) \times 100 \approx 52.7\% \]

### Final Answer:
The percentage of calcium oxide (CaO) in the mixture is approximately **52.7%**.