`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)`
The equivalent of `H_2O_2` reacted with `Sn^(2+)` is

Options

A. `0.2`
B. `0.3`
C. `0.4`
D. `0.6`

Explanation

To find the equivalent of `H_2O_2` that reacted with `Sn^(2+)`, we will follow these steps:

### Step 1: Calculate the mass of `H_2O_2`
Given that we have `136 g` of a `10%` solution of `H_2O_2`, we can find the mass of `H_2O_2` in the solution.

\[ \text{Mass of } H_2O_2 = \frac{10}{100} \times 136 \, \text{g} = 13.6 \, \text{g} \]

### Step 2: Calculate the number of moles of `H_2O_2`
Next, we need to calculate the number of moles of `H_2O_2`. The molar mass of `H_2O_2` is approximately `34 g/mol`.

\[ \text{Moles of } H_2O_2 = \frac{\text{Mass}}{\text{Molar mass}} = \frac{13.6 \, \text{g}}{34 \, \text{g/mol}} \approx 0.4 \, \text{mol} \]

### Step 3: Determine the equivalent of `H_2O_2`
From the reaction provided, we can see that `1 mol` of `H_2O_2` reacts with `2 mol` of electrons (or `Sn^(2+)`). Therefore, the number of equivalents of `H_2O_2` can be calculated as follows:

\[ \text{Equivalents of } H_2O_2 = \text{Moles of } H_2O_2 \times n \text{ factor} \]

Where the n factor for `H_2O_2` is `2` (since it can donate `2` electrons).

\[ \text{Equivalents of } H_2O_2 = 0.4 \, \text{mol} \times 2 = 0.8 \, \text{equivalents} \]

### Step 4: Calculate the equivalents of `Sn^(2+)`
Now, we need to calculate the equivalents of `Sn^(2+)`. The molarity of `Sn^(2+)` is given as `3 M` and the volume is `100 mL` (which is `0.1 L`).

Using the formula for equivalents:

\[ \text{Equivalents of } Sn^{2+} = \text{Normality} \times \text{Volume in L} \]

Normality can be calculated as:

\[ \text{Normality} = \text{Molarity} \times n \text{ factor} = 3 \, \text{M} \times 2 = 6 \, \text{N} \]

Now, substituting the values:

\[ \text{Equivalents of } Sn^{2+} = 6 \, \text{N} \times 0.1 \, \text{L} = 0.6 \, \text{equivalents} \]

### Step 5: Conclusion
Since the equivalents of `H_2O_2` reacted with `Sn^(2+)` is equal to the equivalents of `Sn^(2+)`, we conclude that:

\[ \text{Equivalents of } H_2O_2 \text{ reacted with } Sn^{2+} = 0.6 \, \text{equivalents} \]