1.44 gram of Titanium (Ti) reacted with excess of `O_(2)` and produced x gram of non-stoichiometric compound Ti `._(0.44)O`. The value of x will be :[Ti = 48]

Options

A. 1.44
B. 2.58
C. 1.77
D. None of these

Answer

To solve the problem, we need to find the mass \( x \) of the non-stoichiometric compound \( \text{Ti}_{0.44}\text{O} \) produced when 1.44 grams of Titanium (Ti) reacts with excess oxygen.

### Step-by-Step Solution:

1. **Determine the number of moles of Titanium (Ti)**:
- The molar mass of Titanium (Ti) is given as 48 g/mol.
- Number of moles of Titanium:
\[ \text{Number of moles of Ti} = \frac{\text{mass of Ti}}{\text{molar mass of Ti}} = \frac{1.44 \text{ g}}{48 \text{ g/mol}} = 0.03 \text{ moles} \]

2. **Write the reaction**:
- The reaction can be represented as:
\[ \text{Ti} + \text{O}_2 \rightarrow \text{Ti}_{0.44}\text{O} \]

3. **Calculate the molar mass of \( \text{Ti}_{0.44}\text{O} \)**:
- The molar mass of \( \text{Ti}_{0.44}\text{O} \) can be calculated as follows:
- Molar mass of \( \text{Ti}_{0.44} = 0.44 \times 48 \text{ g/mol} = 21.12 \text{ g/mol} \)
- Molar mass of oxygen (O) = 16 g/mol
- Therefore, molar mass of \( \text{Ti}_{0.44}\text{O} = 21.12 \text{ g/mol} + 16 \text{ g/mol} = 37.12 \text{ g/mol} \)

4. **Set up the relationship using the number of moles**:
- Let \( x \) be the mass of \( \text{Ti}_{0.44}\text{O} \) produced.
- Number of moles of \( \text{Ti}_{0.44}\text{O} \):
\[ \text{Number of moles of } \text{Ti}_{0.44}\text{O} = \frac{x}{37.12} \]

5. **Use the conservation of moles**:
- According to the principle of atomic conservation, the number of moles of Titanium in the reactants must equal the number of moles of Titanium in the products:
\[ \text{Number of moles of Ti} = \text{Number of moles of } \text{Ti}_{0.44}\text{O} \times 0.44 \]
- Therefore:
\[ 0.03 = \frac{0.44x}{37.12} \]

6. **Solve for \( x \)**:
- Rearranging the equation:
\[ x = \frac{0.03 \times 37.12}{0.44} \]
- Calculating \( x \):
\[ x = \frac{1.1136}{0.44} \approx 2.53 \text{ g} \]

7. **Final result**:
- The value of \( x \) is approximately 2.53 grams.

### Conclusion:
The value of \( x \) will be approximately **2.53 grams**.