Calculate the percentage of `BaO` in `29.0 g` mixture of `BaO` and `CaO` which just reacts with `100.8 mL` of `6.0M HCl`.

Explanation

To solve the problem of calculating the percentage of BaO in a 29.0 g mixture of BaO and CaO that reacts with 100.8 mL of 6.0 M HCl, we can follow these steps:

### Step-by-Step Solution:

1. **Calculate the moles of HCl:**
- Use the formula:
\[ \text{Moles of HCl} = \text{Molarity} \times \text{Volume in Liters} \]
- Convert the volume from mL to L:
\[ 100.8 \, \text{mL} = \frac{100.8}{1000} = 0.1008 \, \text{L} \]
- Now calculate the moles:
\[ \text{Moles of HCl} = 6.0 \, \text{M} \times 0.1008 \, \text{L} = 0.6048 \, \text{moles} \]

2. **Set up the equations for the moles of BaO and CaO:**
- Let \( x \) be the mass of BaO in grams.
- The mass of CaO will then be \( 29.0 - x \) grams.
- Molar mass of BaO = 153.33 g/mol.
- Molar mass of CaO = 56.0 g/mol.
- Moles of BaO:
\[ \text{Moles of BaO} = \frac{x}{153.33} \]
- Moles of CaO:
\[ \text{Moles of CaO} = \frac{29.0 - x}{56.0} \]

3. **Write the reaction equations:**
- The balanced reactions are:
- For BaO:
\[ \text{BaO} + 2 \text{HCl} \rightarrow \text{BaCl}_2 + \text{H}_2\text{O} \]
- For CaO:
\[ \text{CaO} + 2 \text{HCl} \rightarrow \text{CaCl}_2 + \text{H}_2\text{O} \]
- Therefore, the total moles of HCl used is:
\[ 2 \left( \frac{x}{153.33} + \frac{29.0 - x}{56.0} \right) = 0.6048 \]

4. **Set up the equation:**
- Rearranging gives:
\[ \frac{x}{153.33} + \frac{29.0 - x}{56.0} = \frac{0.6048}{2} = 0.3024 \]

5. **Multiply through by the least common multiple (LCM) to eliminate fractions:**
- The LCM of 153.33 and 56 is 8596.48. Multiplying through by this gives:
\[ 56x + 153.33(29.0 - x) = 0.3024 \times 8596.48 \]
- Simplifying this will yield a linear equation in \( x \).

6. **Solve for \( x \):**
- After calculations, you will find:
\[ x \approx 19.0 \, \text{g} \]
- This means the mass of BaO in the mixture is 19.0 g.

7. **Calculate the percentage of BaO in the mixture:**
- Use the formula:
\[ \text{Percentage of BaO} = \left( \frac{\text{mass of BaO}}{\text{mass of mixture}} \right) \times 100 \]
- Substituting the values:
\[ \text{Percentage of BaO} = \left( \frac{19.0}{29.0} \right) \times 100 \approx 65.52\% \]

### Final Answer:
The percentage of BaO in the mixture is approximately **65.5%**.