How many moles of `K_(2)Cr_(2) O_(7)` are reduced by 1 mole of formic acid -

Options

A. `1//3` moles
B. 1 mole
C. `2//3` mole
D. `5//3` mole

Answer

To determine how many moles of \( K_2Cr_2O_7 \) are reduced by 1 mole of formic acid (\( HCOOH \)), we can follow these steps:

### Step-by-Step Solution:

1. **Write the balanced chemical equation**:
The reaction between potassium dichromate (\( K_2Cr_2O_7 \)) and formic acid can be written as:
\[ K_2Cr_2O_7 + 3 HCOOH \rightarrow 2 Cr^{3+} + 3 CO_2 + 3 H_2O \]
This equation shows that 1 mole of \( K_2Cr_2O_7 \) reacts with 3 moles of formic acid.

2. **Identify the stoichiometric relationship**:
From the balanced equation, we see that:
- 1 mole of \( K_2Cr_2O_7 \) is reduced by 3 moles of formic acid.

3. **Calculate the moles of \( K_2Cr_2O_7 \) reduced by 1 mole of formic acid**:
If 3 moles of formic acid reduce 1 mole of \( K_2Cr_2O_7 \), then 1 mole of formic acid will reduce:
\[ \text{Moles of } K_2Cr_2O_7 = \frac{1 \text{ mole of } HCOOH}{3 \text{ moles of } HCOOH} = \frac{1}{3} \text{ moles of } K_2Cr_2O_7 \]

4. **Conclusion**:
Therefore, 1 mole of formic acid reduces \( \frac{1}{3} \) moles of \( K_2Cr_2O_7 \).

### Final Answer:
The number of moles of \( K_2Cr_2O_7 \) reduced by 1 mole of formic acid is \( \frac{1}{3} \).

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