`H_2O_2` is reduced rapidly by `Sn^(2+). H_2O_2` is decomposed slowly at room temperature to yeild `O_2` and `H_2O_2`. `136g` of `10%` by mass of `H_2O_2` in water is treated with `100mL` of `3M Sn^(2+)` and then a mixture is allowed to stand until no further reaction occurs. The reactions involved are:
`2H^(o+)+H_(2)O_(2)+Sn^(2+)toSn^(4+)+2H_(2)`
`2H_(2)O_(2)to2H_(2)O+O_(2)`
Calculate the volume of `O_2` produced at `27^@C` and `1` atm after `H_2O_2` is reacted with `Sn^(2+) and the mixture is allowed to stand.

Options

A. `2.46 L`
B. `4.92 L`
C. `1.23 L`
D. `7.38 L`

Explanation

To solve the problem, we need to calculate the volume of oxygen gas (O₂) produced after the reaction of hydrogen peroxide (H₂O₂) with tin(II) ions (Sn²⁺) and the subsequent decomposition of any remaining H₂O₂. Here’s a step-by-step breakdown of the solution:

### Step 1: Calculate the mass of H₂O₂
Given that we have 136 g of a 10% by mass solution of H₂O₂ in water, we can calculate the mass of H₂O₂ in the solution.

\[ \text{Mass of H₂O₂} = \frac{10}{100} \times 136 \, \text{g} = 13.6 \, \text{g} \]

### Step 2: Calculate the moles of H₂O₂
Next, we need to convert the mass of H₂O₂ to moles. The molar mass of H₂O₂ is approximately 34 g/mol.

\[ \text{Moles of H₂O₂} = \frac{13.6 \, \text{g}}{34 \, \text{g/mol}} = 0.4 \, \text{mol} \]

### Step 3: Calculate the moles of Sn²⁺
We are given a 100 mL solution of 3 M Sn²⁺. We can calculate the moles of Sn²⁺.

\[ \text{Moles of Sn²⁺} = \text{Volume (L)} \times \text{Concentration (mol/L)} = 0.1 \, \text{L} \times 3 \, \text{mol/L} = 0.3 \, \text{mol} \]

### Step 4: Identify the limiting reagent
From the reaction equation:

\[ 2H^+ + H₂O₂ + Sn^{2+} \rightarrow Sn^{4+} + 2H₂ + O₂ \]

We can see that 1 mole of Sn²⁺ reacts with 1 mole of H₂O₂. Since we have 0.3 moles of Sn²⁺ and 0.4 moles of H₂O₂, Sn²⁺ is the limiting reagent.

### Step 5: Calculate the moles of O₂ produced from Sn²⁺
From the stoichiometry of the reaction, 1 mole of Sn²⁺ produces 0.5 moles of O₂. Therefore, the moles of O₂ produced from the reaction with Sn²⁺ is:

\[ \text{Moles of O₂ from Sn²⁺} = 0.3 \, \text{mol} \]

### Step 6: Calculate the volume of O₂ produced from Sn²⁺
Using the ideal gas law, at standard conditions (1 atm and 27°C), the molar volume of a gas is approximately 24.4 L/mol.

\[ \text{Volume of O₂ from Sn²⁺} = \text{Moles of O₂} \times \text{Molar Volume} = 0.3 \, \text{mol} \times 24.4 \, \text{L/mol} = 7.32 \, \text{L} \]

### Step 7: Calculate the remaining moles of H₂O₂
After the reaction, the remaining moles of H₂O₂ can be calculated:

\[ \text{Remaining moles of H₂O₂} = 0.4 \, \text{mol} - 0.3 \, \text{mol} = 0.1 \, \text{mol} \]

### Step 8: Calculate the O₂ produced from the decomposition of remaining H₂O₂
The decomposition reaction of H₂O₂ is:

\[ 2H₂O₂ \rightarrow 2H₂O + O₂ \]

From this, we see that 2 moles of H₂O₂ produce 1 mole of O₂. Therefore, 0.1 moles of H₂O₂ will produce:

\[ \text{Moles of O₂ from decomposition} = \frac{0.1}{2} = 0.05 \, \text{mol} \]

### Step 9: Calculate the volume of O₂ from decomposition
Using the molar volume again:

\[ \text{Volume of O₂ from decomposition} = 0.05 \, \text{mol} \times 24.4 \, \text{L/mol} = 1.22 \, \text{L} \]

### Step 10: Calculate the total volume of O₂ produced
Finally, we can add the volumes of O₂ produced from both reactions:

\[ \text{Total Volume of O₂} = 7.32 \, \text{L} + 1.22 \, \text{L} = 8.54 \, \text{L} \]

### Final Answer
The total volume of O₂ produced is approximately **8.54 L**.

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