Metallic tin in the presence of HCl is oxidized by `K_(2)Cr_(2)O_(7)` to stannic chloride, `SnCl_(4)` . What volume of deci-normal dichromate solution would be reduced by 1 g of tin.

Answer

To solve the problem, we need to determine the volume of deci-normal (0.1 N) dichromate solution that would be reduced by 1 g of tin (Sn) when oxidized to stannic chloride (SnCl4). Here is a step-by-step solution:

### Step 1: Determine the equivalent weight of tin (Sn)
The equivalent weight of a substance is calculated using the formula:
\[ \text{Equivalent Weight} = \frac{\text{Molar Mass}}{n} \]
where \( n \) is the number of electrons exchanged during the reaction.

For tin (Sn):
- Molar mass of Sn = 118.71 g/mol
- In this reaction, Sn is oxidized from 0 to +4, meaning it loses 4 electrons. Thus, \( n = 4 \).

Calculating the equivalent weight:
\[ \text{Equivalent Weight of Sn} = \frac{118.71 \, \text{g/mol}}{4} = 29.6775 \, \text{g/equiv} \]

### Step 2: Calculate the number of equivalents of tin in 1 g
Using the formula:
\[ \text{Number of Equivalents} = \frac{\text{Mass}}{\text{Equivalent Weight}} \]
Substituting the values:
\[ \text{Number of Equivalents of Sn} = \frac{1 \, \text{g}}{29.6775 \, \text{g/equiv}} \approx 0.0337 \, \text{equiv} \]

### Step 3: Relate the equivalents of tin to the equivalents of K2Cr2O7
In the redox reaction, the equivalents of tin will equal the equivalents of K2Cr2O7:
\[ \text{Equivalents of Sn} = \text{Equivalents of K2Cr2O7} \]
Thus, the equivalents of K2Cr2O7 required will also be approximately 0.0337 equiv.

### Step 4: Calculate the volume of K2Cr2O7 solution needed
Using the formula:
\[ \text{Equivalents} = \text{Normality} \times \text{Volume} \]
Rearranging gives:
\[ \text{Volume} = \frac{\text{Equivalents}}{\text{Normality}} \]
Substituting the values:
- Normality of K2Cr2O7 = 0.1 N
- Equivalents of K2Cr2O7 = 0.0337 equiv

Calculating the volume:
\[ \text{Volume} = \frac{0.0337 \, \text{equiv}}{0.1 \, \text{N}} = 0.337 \, \text{L} = 337 \, \text{mL} \]

### Final Answer
The volume of deci-normal dichromate solution that would be reduced by 1 g of tin is approximately **337 mL**.

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