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)`1.4`
B. (b)`3.0`
C. (c )`2.8`
D. (d)`4.4`

Explanation

To solve the problem, we need to determine 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. **Calculate the number of moles of formic acid (HCOOH):**
- Given mass of formic acid = 2.3 g
- Molecular weight of formic acid (HCOOH) = 12 (C) + 1 (H) + 16 (O) + 1 (H) = 46 g/mol
- Number of moles of formic acid = mass / molecular weight = 2.3 g / 46 g/mol = 0.05 moles

2. **Calculate the number of moles of oxalic acid (C2H2O4):**
- Given mass of oxalic acid = 4.5 g
- Molecular weight of oxalic acid (C2H2O4) = 2*12 (C) + 2*1 (H) + 4*16 (O) = 90 g/mol
- Number of moles of oxalic acid = mass / molecular weight = 4.5 g / 90 g/mol = 0.05 moles

3. **Determine the products formed from the reactions:**
- **From formic acid:**
- 1 mole of formic acid produces 1 mole of CO (carbon monoxide) and 1 mole of H2O (water).
- Therefore, from 0.05 moles of formic acid, we get:
- 0.05 moles of CO
- 0.05 moles of H2O
- **From oxalic acid:**
- 1 mole of oxalic acid produces 1 mole of CO2 (carbon dioxide), 1 mole of CO, and 1 mole of H2O.
- Therefore, from 0.05 moles of oxalic acid, we get:
- 0.05 moles of CO2
- 0.05 moles of CO
- 0.05 moles of H2O

4. **Calculate the total moles of gases before passing through KOH:**
- Total moles of CO = 0.05 (from formic acid) + 0.05 (from oxalic acid) = 0.1 moles
- Total moles of CO2 = 0.05 moles
- Total moles of H2O = 0.05 moles
- Thus, total moles of gas = 0.1 (CO) + 0.05 (CO2) + 0.05 (H2O) = 0.2 moles

5. **Effect of passing through KOH pellets:**
- KOH absorbs CO2. Therefore, the moles of CO2 will be removed from the gaseous mixture.
- Remaining gases after passing through KOH = 0.1 moles of CO + (0.05 - 0.05) moles of CO2 + 0.05 moles of H2O = 0.1 moles of CO + 0.05 moles of H2O

6. **Calculate the weight of the remaining product:**
- Remaining gases = 0.1 moles of CO + 0.05 moles of H2O
- Molecular weight of CO = 12 (C) + 16 (O) = 28 g/mol
- Molecular weight of H2O = 2 (H) + 16 (O) = 18 g/mol
- Total weight of remaining gases = (0.1 moles * 28 g/mol) + (0.05 moles * 18 g/mol)
- Total weight = 2.8 g (from CO) + 0.9 g (from H2O) = 3.7 g

### Final Answer:
The weight of the remaining product at STP will be **3.7 g**.