A mixture of 2.3 g formic acid and 4.5 g oxalic acid is treated with conc. `H_(2)SO_(4)`. The evolved gaseous mixture is passed through KOH pellets. Weight (in g) of the remaining product at STP will be

Options

A. (a) 2.8
B. (b) 3
C. (c) 1.4
D. (d) 4.4

Explanation

To solve the problem, we need to calculate the weight of the remaining product after treating a mixture of formic acid and oxalic acid with concentrated sulfuric acid and then passing the evolved gases through KOH pellets.

### Step-by-Step Solution:

1. **Identify the acids and their weights:**
- Formic acid (HCOOH): 2.3 g
- Oxalic acid (C2H2O4): 4.5 g

2. **Calculate the molecular weight of formic acid (HCOOH):**
- H: 1 g × 2 = 2 g
- C: 12 g × 1 = 12 g
- O: 16 g × 2 = 32 g
- Total = 2 + 12 + 32 = 46 g/mol

3. **Calculate the moles of formic acid:**
\[ \text{Moles of HCOOH} = \frac{\text{Given weight}}{\text{Molecular weight}} = \frac{2.3 \text{ g}}{46 \text{ g/mol}} = \frac{2.3}{46} \approx 0.050 \text{ moles} \]

4. **Reaction of formic acid with concentrated H2SO4:**
- The reaction produces carbon monoxide (CO) and water (H2O).
- Moles of CO produced = Moles of HCOOH reacted = 0.050 moles.

5. **Calculate the molecular weight of oxalic acid (C2H2O4):**
- C: 12 g × 2 = 24 g
- H: 1 g × 2 = 2 g
- O: 16 g × 4 = 64 g
- Total = 24 + 2 + 64 = 90 g/mol

6. **Calculate the moles of oxalic acid:**
\[ \text{Moles of C2H2O4} = \frac{4.5 \text{ g}}{90 \text{ g/mol}} = \frac{4.5}{90} = 0.050 \text{ moles} \]

7. **Reaction of oxalic acid with concentrated H2SO4:**
- The reaction produces carbon monoxide (CO), carbon dioxide (CO2), and water (H2O).
- Moles of CO produced = Moles of C2H2O4 reacted = 0.050 moles.
- Moles of CO2 produced = 0.050 moles.

8. **Total moles of gases produced:**
- From formic acid: 0.050 moles of CO
- From oxalic acid: 0.050 moles of CO + 0.050 moles of CO2
- Total moles of CO = 0.050 + 0.050 = 0.100 moles of CO
- Total moles of CO2 = 0.050 moles of CO2

9. **Passing the gas mixture through KOH pellets:**
- KOH absorbs CO2, leaving CO in the remaining gas.
- Remaining gas = 0.100 moles of CO.

10. **Calculate the mass of the remaining CO:**
- Molar mass of CO = 12 g (C) + 16 g (O) = 28 g/mol
\[ \text{Mass of remaining CO} = \text{Moles} \times \text{Molar mass} = 0.100 \text{ moles} \times 28 \text{ g/mol} = 2.8 \text{ g} \]

### Final Answer:
The weight of the remaining product at STP will be **2.8 g**.