In the reaction `CrO_(5) + H_(2)SO_(4)rarr Cr_(2)(sO_(4))_(3)+H_(2)O+O_(2)`, one mole of `CrO_(5)` will liberate how many molesof `O_(2)` :

Options

A. `5//2`
B. `5//4`
C. `9//2`
D. `7//4`

Explanation

To determine how many moles of \( O_2 \) are liberated from one mole of \( CrO_5 \) in the reaction:

\[ CrO_5 + H_2SO_4 \rightarrow Cr_2(SO_4)_3 + H_2O + O_2 \]

we need to follow these steps:

### Step 1: Write the balanced equation
The first step is to balance the chemical equation. The balanced equation for the reaction is:

\[ 4 CrO_5 + 6 H_2SO_4 \rightarrow 2 Cr_2(SO_4)_3 + 6 H_2O + 7 O_2 \]

### Step 2: Analyze the stoichiometry
From the balanced equation, we can see the stoichiometric relationship between \( CrO_5 \) and \( O_2 \).

### Step 3: Determine moles of \( O_2 \) produced per mole of \( CrO_5 \)
According to the balanced equation:
- 4 moles of \( CrO_5 \) produce 7 moles of \( O_2 \).

To find out how many moles of \( O_2 \) are produced by 1 mole of \( CrO_5 \), we can set up a proportion:

\[ \text{Moles of } O_2 = \frac{7 \text{ moles of } O_2}{4 \text{ moles of } CrO_5} \]

### Step 4: Calculate the moles of \( O_2 \) for 1 mole of \( CrO_5 \)
Now, we can calculate the moles of \( O_2 \) produced from 1 mole of \( CrO_5 \):

\[ \text{Moles of } O_2 = \frac{7}{4} = 1.75 \]

### Conclusion
Thus, 1 mole of \( CrO_5 \) will liberate \( \frac{7}{4} \) moles of \( O_2 \).

### Final Answer
The final answer is \( \frac{7}{4} \) moles of \( O_2 \).

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