3.4 gm of `H_(2)O_(2)` decomposes, the weight of oxygen liberated from it is

Options

A. 1.6 gm
B. 2.24 gm
C. 1.16 gm
D. 3.2 gm

Explanation

To find the weight of oxygen liberated from the decomposition of 3.4 grams of hydrogen peroxide (H₂O₂), we can follow these steps:

### Step 1: Write the decomposition reaction of H₂O₂
The decomposition of hydrogen peroxide can be represented by the following balanced chemical equation:
\[ 2 H_2O_2 \rightarrow 2 H_2O + O_2 \]

### Step 2: Calculate the molar mass of H₂O₂
The molar mass of H₂O₂ can be calculated as follows:
- Hydrogen (H) has an atomic mass of approximately 1 g/mol, and there are 2 hydrogen atoms: \(2 \times 1 = 2 \, \text{g/mol}\)
- Oxygen (O) has an atomic mass of approximately 16 g/mol, and there are 2 oxygen atoms: \(2 \times 16 = 32 \, \text{g/mol}\)

Thus, the molar mass of H₂O₂ is:
\[ 2 + 32 = 34 \, \text{g/mol} \]

### Step 3: Determine the moles of H₂O₂ decomposed
To find the number of moles of H₂O₂ in 3.4 grams, we use the formula:
\[ \text{Moles of H₂O₂} = \frac{\text{mass}}{\text{molar mass}} = \frac{3.4 \, \text{g}}{34 \, \text{g/mol}} = 0.1 \, \text{mol} \]

### Step 4: Use the stoichiometry of the reaction to find moles of O₂ produced
From the balanced equation, we see that 2 moles of H₂O₂ produce 1 mole of O₂. Therefore, the moles of O₂ produced from 0.1 moles of H₂O₂ can be calculated as follows:
\[ \text{Moles of O₂} = \frac{0.1 \, \text{mol H₂O₂}}{2} = 0.05 \, \text{mol O₂} \]

### Step 5: Calculate the mass of O₂ produced
Now, we can find the mass of the oxygen produced using its molar mass (O₂ has a molar mass of 32 g/mol):
\[ \text{Mass of O₂} = \text{moles} \times \text{molar mass} = 0.05 \, \text{mol} \times 32 \, \text{g/mol} = 1.6 \, \text{g} \]

### Final Answer
The weight of oxygen liberated from the decomposition of 3.4 grams of H₂O₂ is **1.6 grams**.

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