The volume of oxygen liberated from `0.68g` of `H_(2)O_(2)` is

Options

A. 112 ml
B. 224 ml
C. 56 ml
D. 336 ml

Answer

To find the volume of oxygen liberated from 0.68 g of hydrogen peroxide (H₂O₂), we can follow these steps:

### Step 1: Write the balanced chemical equation for the decomposition of hydrogen peroxide.

The decomposition of hydrogen peroxide can be represented by the following balanced equation:

\[ 2 H_2O_2 \rightarrow 2 H_2O + O_2 \]

This equation tells us that 2 moles of hydrogen peroxide decompose to produce 1 mole of oxygen gas.

### Step 2: Calculate the molar mass of hydrogen peroxide (H₂O₂).

The molar mass of H₂O₂ can be calculated as follows:

- Hydrogen (H): 1 g/mol × 2 = 2 g/mol
- Oxygen (O): 16 g/mol × 2 = 32 g/mol

Thus, the molar mass of H₂O₂ is:

\[ 2 + 32 = 34 \text{ g/mol} \]

### Step 3: Calculate the number of moles of hydrogen peroxide in 0.68 g.

Using the molar mass, we can find the number of moles of H₂O₂:

\[ \text{Number of moles of } H_2O_2 = \frac{\text{mass}}{\text{molar mass}} = \frac{0.68 \text{ g}}{34 \text{ g/mol}} = 0.02 \text{ moles} \]

### Step 4: Determine the moles of oxygen produced.

From the balanced equation, we know that 2 moles of H₂O₂ produce 1 mole of O₂. Therefore, the moles of O₂ produced from 0.02 moles of H₂O₂ can be calculated as follows:

\[ \text{Moles of } O_2 = \frac{0.02 \text{ moles of } H_2O_2}{2} = 0.01 \text{ moles of } O_2 \]

### Step 5: Calculate the volume of oxygen at NTP.

At Normal Temperature and Pressure (NTP), 1 mole of any gas occupies 22.4 liters. Therefore, the volume of 0.01 moles of O₂ can be calculated as:

\[ \text{Volume of } O_2 = 0.01 \text{ moles} \times 22.4 \text{ L/mole} = 0.224 \text{ L} \]

### Step 6: Convert the volume from liters to milliliters.

To convert liters to milliliters, we multiply by 1000:

\[ 0.224 \text{ L} \times 1000 = 224 \text{ mL} \]

### Conclusion

The volume of oxygen liberated from 0.68 g of hydrogen peroxide is **224 mL**.

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