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. 2.8
B. `3.0`
C. 1.4
D. 4.4

Explanation

To solve the problem step by step, we will follow the chemical reactions and calculations involved in the process.

### Step 1: Determine the moles of formic acid and oxalic acid.

**Formic Acid (HCOOH):**
- Given mass = 2.3 g
- Molecular weight = 1 (H) + 12 (C) + 16 (O) + 1 (H) = 46 g/mol
- Moles of formic acid = Mass / Molecular weight = 2.3 g / 46 g/mol = 0.05 moles

**Oxalic Acid (C2H2O4):**
- Given mass = 4.5 g
- Molecular weight = 2(12) + 4(1) + 4(16) = 90 g/mol
- Moles of oxalic acid = Mass / Molecular weight = 4.5 g / 90 g/mol = 0.05 moles

### Step 2: Write the reactions that occur when treated with concentrated H2SO4.

1. **For Formic Acid:**
\[ \text{HCOOH} \xrightarrow{\text{conc. } H_2SO_4} \text{CO} + \text{H}_2O \]
- From 1 mole of formic acid, 1 mole of CO is produced.

2. **For Oxalic Acid:**
\[ \text{C}_2\text{H}_2\text{O}_4 \xrightarrow{\text{conc. } H_2SO_4} \text{CO} + \text{CO}_2 + \text{H}_2O \]
- From 1 mole of oxalic acid, 1 mole of CO and 1 mole of CO2 are produced.

### Step 3: Calculate the total moles of gases produced.

- Moles of CO from formic acid = 0.05 moles
- Moles of CO from oxalic acid = 0.05 moles
- Moles of CO2 from oxalic acid = 0.05 moles

**Total moles of gases produced:**
\[ \text{Total CO} = 0.05 + 0.05 = 0.1 \text{ moles} \]
\[ \text{Total CO}_2 = 0.05 \text{ moles} \]

### Step 4: Determine the effect of passing the gas mixture through KOH.

- KOH absorbs CO2. Thus, the moles of CO2 absorbed = 0.05 moles.
- Remaining gas after KOH treatment = Total CO - CO2 absorbed = 0.1 moles of CO + 0 moles of CO2 = 0.1 moles of CO.

### Step 5: Calculate the weight of the remaining product at STP.

- Moles of remaining gas (CO) = 0.1 moles
- Molecular weight of CO = 12 (C) + 16 (O) = 28 g/mol
- Weight of remaining product = Moles × Molecular weight = 0.1 moles × 28 g/mol = 2.8 g

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

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