Element X reacts with oxygen to produce a pure sample of `X_2 O_3`. In a experiment, it is found that 1.00 g of X produces 1.16g of `X_2 O_3`. What will be the atomic mass of X.

Options

A. 67 g/mole
B. 100.2 g/mole
C. 125 g/mole
D. 150 g/mole

Answer

To find the atomic mass of element X that reacts with oxygen to form `X2O3`, we can follow these steps:

### Step 1: Write the balanced chemical equation
The reaction can be represented as:
\[ 4X + 3O_2 \rightarrow 2X_2O_3 \]
This indicates that 2 moles of `X2O3` are produced from 4 moles of X.

### Step 2: Determine the molar mass of `X2O3`
The molar mass of `X2O3` can be calculated as follows:
- Molar mass of `X2O3` = 2 * (atomic mass of X) + 3 * (molar mass of O)
- Molar mass of O = 16 g/mol
- Therefore, Molar mass of `X2O3` = \(2X + 3 \times 16 = 2X + 48\) g/mol

### Step 3: Calculate the number of moles of `X` and `X2O3`
Given:
- Mass of X = 1.00 g
- Mass of `X2O3` = 1.16 g

**Number of moles of X:**
\[ \text{Number of moles of X} = \frac{\text{mass of X}}{\text{molar mass of X}} = \frac{1.00}{X} \]

**Number of moles of `X2O3`:**
\[ \text{Number of moles of } X2O3 = \frac{\text{mass of } X2O3}{\text{molar mass of } X2O3} = \frac{1.16}{(2X + 48)} \]

### Step 4: Relate the moles of X and `X2O3`
From the balanced equation, we know:
\[ \text{Moles of X} = 2 \times \text{Moles of } X2O3 \]
Thus,
\[ \frac{1.00}{X} = 2 \times \frac{1.16}{(2X + 48)} \]

### Step 5: Cross-multiply and simplify
Cross-multiplying gives:
\[ 1.00 \times (2X + 48) = 2 \times 1.16 \times X \]
\[ 2X + 48 = 2.32X \]

### Step 6: Solve for X
Rearranging the equation:
\[ 48 = 2.32X - 2X \]
\[ 48 = 0.32X \]
\[ X = \frac{48}{0.32} = 150 \text{ g/mol} \]

### Conclusion
The atomic mass of element X is **150 g/mol**.