During the decomposition of `H_(2)O_(2)` to give oxygen, `48 g O_(2)` is formed per minute at a certain point of time. The rate of formation of water at this point is

Options

A. `0.75 mol min^(-1)`
B. `1.5 mol min^(-1)`
C. `2.25 mol min^(-1)`
D. `3.0 mol min^(-1)`

Answer

To solve the problem, we need to determine the rate of formation of water during the decomposition of hydrogen peroxide (H₂O₂) into oxygen (O₂) and water (H₂O).

### Step-by-Step Solution:

1. **Write the Balanced Chemical Equation**:
The decomposition of hydrogen peroxide can be represented by the following balanced equation:
\[ 2 \, \text{H}_2\text{O}_2 \rightarrow 2 \, \text{H}_2\text{O} + \text{O}_2 \]

2. **Identify the Molar Mass of Oxygen**:
The molar mass of O₂ (oxygen) is approximately 32 g/mol.

3. **Calculate the Rate of Formation of Moles of Oxygen**:
Given that 48 g of O₂ is formed per minute, we can convert this mass into moles:
\[ \text{Moles of O}_2 = \frac{48 \, \text{g}}{32 \, \text{g/mol}} = 1.5 \, \text{moles/min} \]

4. **Relate the Rate of Formation of Water to Oxygen**:
From the balanced equation, we can see that 1 mole of O₂ is produced for every 2 moles of H₂O formed. Therefore, the rate of formation of water (H₂O) is twice the rate of formation of oxygen (O₂):
\[ \text{Rate of formation of H}_2\text{O} = 2 \times \text{Rate of formation of O}_2 \]

5. **Calculate the Rate of Formation of Water**:
Substituting the rate of formation of O₂:
\[ \text{Rate of formation of H}_2\text{O} = 2 \times 1.5 \, \text{moles/min} = 3 \, \text{moles/min} \]

### Final Answer:
The rate of formation of water at this point is **3 moles/min**.

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