The following reaction takes places in basic medium:
`underset(("chromate ion"))(CrO_(4)^(2-))+underset(("Stannite ion"))(HSnO_(2)^(ɵ))+H_(2)Otounderset(("stannate ion"))(HSnO_(3)^(ɵ))+underset(("Chromite ion"))(CrO_(2)^(ɵ)+overset(ɵ)(O)H)`
If 400 " mL of " `(M)/(5)` chromate ion react with 500 " mL of " `(M)/(4)` stannite ion, then which of the following statements are correct?

Options

A. Chromate ion `CrO_(4)^(2-)` is the limiting reagent.
B. Stannite ion, `HSnO_(2)^(ɵ)` is the limiting reagent.
C. At the end of reaction concentration of `CrO_(2)^(ɵ)` ions `approx0.08M`
D. At the end of reaction concentration of `HSnO_(3)^(ɵ)`
`approx0.13M`

Explanation

To solve the problem, we need to analyze the given reaction and calculate the equivalents of the reactants involved. Let's break it down step by step.

### Step 1: Write down the balanced reaction
The reaction given is:
\[ \text{CrO}_4^{2-} + \text{HSnO}_2^{-} + \text{H}_2\text{O} \rightarrow \text{HSnO}_3^{-} + \text{CrO}_2^{-} + \text{OH}^- \]

### Step 2: Determine the oxidation states
- For **Chromium (Cr)** in \(\text{CrO}_4^{2-}\):
- Oxidation state = +6
- For **Chromium (Cr)** in \(\text{CrO}_2^{-}\):
- Oxidation state = +3
- Thus, Chromium is reduced from +6 to +3.

- For **Tin (Sn)** in \(\text{HSnO}_2^{-}\):
- Let the oxidation state of Sn be \(x\):
- \(x + 2(-2) + 1 = -1\) (from the charge of the ion)
- \(x - 4 + 1 = -1\)
- \(x - 3 = -1 \Rightarrow x = +2\)

- For **Tin (Sn)** in \(\text{HSnO}_3^{-}\):
- Let the oxidation state of Sn be \(y\):
- \(y + 3(-2) + 1 = -1\)
- \(y - 6 + 1 = -1\)
- \(y - 5 = -1 \Rightarrow y = +4\)

Thus, Tin is oxidized from +2 to +4.

### Step 3: Calculate the milliequivalents of each reactant
1. **Chromate ion (\(\text{CrO}_4^{2-}\))**:
- Molarity = \( \frac{1}{5} \, M \)
- Volume = \( 400 \, mL = 0.4 \, L \)
- Milliequivalents = \( \text{Molarity} \times \text{Volume} \times \text{n-factor} \)
- n-factor for \(\text{CrO}_4^{2-}\) (change in oxidation state from +6 to +3) = 3
- Milliequivalents = \( \frac{1}{5} \times 0.4 \times 3 = 240 \, \text{meq} \)

2. **Stannite ion (\(\text{HSnO}_2^{-}\))**:
- Molarity = \( \frac{1}{4} \, M \)
- Volume = \( 500 \, mL = 0.5 \, L \)
- n-factor for \(\text{HSnO}_2^{-}\) (change in oxidation state from +2 to +4) = 2
- Milliequivalents = \( \frac{1}{4} \times 0.5 \times 2 = 250 \, \text{meq} \)

### Step 4: Determine the limiting reagent
- From the calculations:
- \(\text{CrO}_4^{2-}\) = 240 meq
- \(\text{HSnO}_2^{-}\) = 250 meq

Since \(\text{CrO}_4^{2-}\) has fewer milliequivalents, it is the limiting reagent.

### Step 5: Analyze the reaction outcome
- The reaction will consume all of the \(\text{CrO}_4^{2-}\) (240 meq), and it will require:
- \( \frac{240 \, \text{meq}}{3} = 80 \, \text{meq} \) of \(\text{HSnO}_2^{-}\).
- Remaining \(\text{HSnO}_2^{-}\):
- Initial = 250 meq
- Consumed = 80 meq
- Remaining = \( 250 - 80 = 170 \, \text{meq} \)

### Conclusion
- The limiting reagent is \(\text{CrO}_4^{2-}\).
- After the reaction, there will be excess \(\text{HSnO}_2^{-}\).

### Final Statements
Based on the calculations, the correct statements regarding the reaction are:
- The limiting reagent is \(\text{CrO}_4^{2-}\).
- There will be excess \(\text{HSnO}_2^{-}\) remaining after the reaction.