`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`. `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 step by step, we will follow the reactions and calculations as outlined in the video transcript.

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

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

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

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

### Step 3: Calculate the equivalents of H₂O₂ and Sn²⁺
The reaction shows that 1 mole of H₂O₂ reacts with 1 mole of Sn²⁺. However, the n-factor for H₂O₂ in this reaction is 2 because it changes from an oxidation state of -1 to -2 (0 to -1 for hydrogen).

Thus, the equivalents of H₂O₂ are:
\[ \text{Equivalents of H₂O₂} = \text{Moles of H₂O₂} \times \text{n-factor} = 0.4 \, \text{mol} \times 2 = 0.8 \, \text{equivalents} \]

For Sn²⁺, we have:
- Molarity = 3 M
- Volume = 100 mL = 0.1 L
- n-factor = 2 (as it goes from +2 to +4)

Calculating the equivalents of Sn²⁺:
\[ \text{Equivalents of Sn²⁺} = \text{Molarity} \times \text{Volume} \times \text{n-factor} = 3 \, \text{mol/L} \times 0.1 \, \text{L} \times 2 = 0.6 \, \text{equivalents} \]

### Step 4: Determine the limiting reagent
Since the reaction consumes H₂O₂ and Sn²⁺, we need to determine which reactant is the limiting reagent.

- H₂O₂: 0.8 equivalents available
- Sn²⁺: 0.6 equivalents available

Since Sn²⁺ has fewer equivalents, it is the limiting reagent.

### Step 5: Calculate the moles of H₂O₂ that reacted
Using the limiting reagent (Sn²⁺), we can find out how many moles of H₂O₂ reacted. Since 1 equivalent of Sn²⁺ reacts with 1 equivalent of H₂O₂, the amount of H₂O₂ that reacted is equal to the equivalents of Sn²⁺.

\[ \text{Moles of H₂O₂ reacted} = 0.6 \, \text{equivalents} \div 2 = 0.3 \, \text{mol} \]

### Step 6: Calculate the moles of H₂O₂ left
Now, we can find the moles of H₂O₂ left after the reaction:

\[ \text{Moles of H₂O₂ left} = \text{Initial moles of H₂O₂} - \text{Moles of H₂O₂ reacted} \]
\[ \text{Moles of H₂O₂ left} = 0.4 \, \text{mol} - 0.3 \, \text{mol} = 0.1 \, \text{mol} \]

### Step 7: Calculate the moles of O₂ produced
According to the decomposition reaction of H₂O₂:
\[ 2 \, \text{H₂O₂} \rightarrow 2 \, \text{H₂O} + \text{O₂} \]
From the stoichiometry, 2 moles of H₂O₂ produce 1 mole of O₂. Therefore, 0.1 moles of H₂O₂ will produce:

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

### Step 8: Calculate the volume of O₂ produced
Using the ideal gas law \(PV = nRT\), we can calculate the volume of O₂ produced at 27°C (300 K) and 1 atm.

\[ V = nRT/P \]
Where:
- \(n = 0.05 \, \text{mol}\)
- \(R = 0.0821 \, \text{L atm/(mol K)}\)
- \(T = 300 \, \text{K}\)
- \(P = 1 \, \text{atm}\)

Substituting the values:

\[ V = \frac{0.05 \, \text{mol} \times 0.0821 \, \text{L atm/(mol K)} \times 300 \, \text{K}}{1 \, \text{atm}} \approx 1.23 \, \text{L} \]

### Final Answer
The volume of O₂ produced is approximately **1.23 liters**.

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