`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)`
The volume strength of `H_2O_2` left after reacting with `Sn^(2+)`

Options

A. `1.12 V`
B. `11.2 V`
C. `2.24 V`
D. `22.4 V`

Explanation

To solve the problem, we will follow these steps:

### 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.

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

### Step 2: Calculate the moles of Sn²⁺
We are given that we have 100 mL of a 3 M Sn²⁺ solution. To find the moles of Sn²⁺, we use the formula:

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

### Step 3: Determine the moles of H₂O₂ that react
From the stoichiometry of the reaction, we know that 1 mole of Sn²⁺ reacts with 1 mole of H₂O₂. Therefore, the moles of H₂O₂ that react will be equal to the moles of Sn²⁺.

\[ \text{Moles of H₂O₂ reacted} = \text{Moles of Sn²⁺} = 0.3 \, \text{mol} \]

### Step 4: Calculate the mass of H₂O₂ that reacted
To find the mass of H₂O₂ that reacted, we use the molar mass of H₂O₂, which is approximately 34 g/mol.

\[ \text{Mass of H₂O₂ reacted} = \text{Moles} \times \text{Molar mass} = 0.3 \, \text{mol} \times 34 \, \text{g/mol} = 10.2 \, \text{g} \]

### Step 5: Calculate the mass of unreacted H₂O₂
Now we can find the mass of H₂O₂ that remains unreacted.

\[ \text{Mass of H₂O₂ unreacted} = \text{Total mass of H₂O₂} - \text{Mass of H₂O₂ reacted} = 13.6 \, \text{g} - 10.2 \, \text{g} = 3.4 \, \text{g} \]

### Step 6: Calculate the moles of unreacted H₂O₂
Now we can calculate the moles of unreacted H₂O₂.

\[ \text{Moles of H₂O₂ unreacted} = \frac{\text{Mass}}{\text{Molar mass}} = \frac{3.4 \, \text{g}}{34 \, \text{g/mol}} = 0.1 \, \text{mol} \]

### Step 7: Calculate the volume strength of H₂O₂
The volume strength of H₂O₂ can be calculated using the molarity of the solution. The molarity of the unreacted solution is:

\[ \text{Molarity of H₂O₂} = \frac{\text{Moles}}{\text{Volume (L)}} = \frac{0.1 \, \text{mol}}{0.1 \, \text{L}} = 1 \, \text{M} \]

The volume strength is given by:

\[ \text{Volume strength} = \text{Molarity} \times 11.2 = 1 \, \text{M} \times 11.2 = 11.2 \]

### Final Answer
The volume strength of H₂O₂ left after reacting with Sn²⁺ is **11.2 volume**.

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