How many moles of `O_(2)` will be liberated by one mole of `CrO_(5)` is the following reaction:
`CrO_(5) + H_(2) SO_(4) rarr Cr_(2) (SO_(4))_(5) + H_(2)O + O_(2)`

Options

A. 4.5
B. 2.5
C. 1.25
D. None

Explanation

To determine how many moles of \( O_2 \) will be liberated by one mole of \( CrO_5 \) in the given reaction, we can follow these steps:

### Step 1: Write the balanced chemical equation
The unbalanced reaction is:
\[ CrO_5 + H_2SO_4 \rightarrow Cr_2(SO_4)_3 + H_2O + O_2 \]

### Step 2: Balance the reaction
To balance the reaction, we need to ensure that the number of atoms of each element is the same on both sides. The balanced equation is:
\[ 2 CrO_5 + 3 H_2SO_4 \rightarrow Cr_2(SO_4)_3 + 3 H_2O + \frac{7}{2} O_2 \]

### Step 3: Determine the mole ratio
From the balanced equation, we can see that:
- 2 moles of \( CrO_5 \) produce \( \frac{7}{2} \) moles of \( O_2 \).

### Step 4: Calculate moles of \( O_2 \) produced by 1 mole of \( CrO_5 \)
To find out how many moles of \( O_2 \) are produced by 1 mole of \( CrO_5 \), we can set up a proportion based on the mole ratio:
\[ \text{If } 2 \text{ moles of } CrO_5 \text{ produce } \frac{7}{2} \text{ moles of } O_2, \]
\[ \text{Then } 1 \text{ mole of } CrO_5 \text{ will produce } \frac{7}{2} \times \frac{1}{2} = \frac{7}{4} \text{ moles of } O_2. \]

### Step 5: Convert to decimal
Calculating \( \frac{7}{4} \):
\[ \frac{7}{4} = 1.75 \text{ moles of } O_2. \]

### Conclusion
Thus, 1 mole of \( CrO_5 \) will liberate **1.75 moles of \( O_2 \)**.