The volume of `O_(2)` liberated from 0.96g of `H_(2)O_(2)` is

Options

A. 224.6 mL
B. 316.2 mL
C. 390.0 mL
D. 112.5 mL

Answer

To find the volume of \( O_2 \) liberated from 0.96 g of \( H_2O_2 \), we can follow these steps:

### Step 1: Write the decomposition reaction of hydrogen peroxide
The decomposition of hydrogen peroxide can be represented by the following chemical equation:
\[ 2 H_2O_2 \rightarrow 2 H_2O + O_2 \]
From this equation, we can see that 2 moles of hydrogen peroxide produce 1 mole of oxygen gas.

### Step 2: Calculate the molar mass of hydrogen peroxide
The molar mass of \( H_2O_2 \) (hydrogen peroxide) is calculated as follows:
- Hydrogen (H) = 1 g/mol, and there are 2 hydrogen atoms.
- Oxygen (O) = 16 g/mol, and there are 2 oxygen atoms.

Thus, the molar mass of \( H_2O_2 \) is:
\[ (2 \times 1) + (2 \times 16) = 2 + 32 = 34 \text{ g/mol} \]

### Step 3: Calculate the number of moles of \( H_2O_2 \) in 0.96 g
Using the molar mass, we can find the number of moles of \( H_2O_2 \):
\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{0.96 \text{ g}}{34 \text{ g/mol}} \approx 0.02824 \text{ moles} \]

### Step 4: Determine the moles of \( O_2 \) produced
From the balanced equation, we know that 2 moles of \( H_2O_2 \) produce 1 mole of \( O_2 \). Therefore, the moles of \( O_2 \) produced from 0.02824 moles of \( H_2O_2 \) can be calculated as follows:
\[ \text{Moles of } O_2 = \frac{0.02824 \text{ moles } H_2O_2}{2} \approx 0.01412 \text{ moles} \]

### Step 5: Calculate the volume of \( O_2 \) produced
At standard temperature and pressure (STP), 1 mole of gas occupies 22,400 mL. Therefore, the volume of \( O_2 \) produced can be calculated as:
\[ \text{Volume of } O_2 = \text{moles of } O_2 \times 22400 \text{ mL/mol} = 0.01412 \text{ moles} \times 22400 \text{ mL/mol} \approx 316.2 \text{ mL} \]

### Conclusion
The volume of \( O_2 \) liberated from 0.96 g of \( H_2O_2 \) is approximately **316.2 mL**.