Question

A mixture of 2.3 g formic acid and 4.5 g oxalic acid is treated with conc. H₂SO₄ . The evolved gaseous mixture is passed through KOH pellets. Weight (in g) of the remaining product at STP will be

• A. 1.4
• B. 2.8
• C. 3.0
• D. 4.4

Answer 1

The Correct Option is B

To solve this problem, we must analyze the chemical reaction and identify the products formed. The question involves the reaction of formic acid (HCOOH) and oxalic acid (H₂C₂O₄) with concentrated sulfuric acid (H₂SO₄). When these acids are reacted with conc. H₂SO₄, CO (carbon monoxide) and CO₂ (carbon dioxide) gases are evolved.

The chemical reactions are as follows:

• For formic acid: \text{HCOOH} \xrightarrow{\text{conc. H}_2\text{SO}_4} \text{CO} + \text{H}_2\text{O}
• For oxalic acid: \text{H}_2\text{C}_2\text{O}_4 \xrightarrow{\text{conc. H}_2\text{SO}_4} \text{CO} + \text{CO}_2 + \text{H}_2\text{O}

The evolved gas mixture of CO and CO₂ is then passed through KOH, which absorbs CO₂ completely, leaving only CO.

Next, we calculate the moles of gases:

• For formic acid (HCOOH) with molar mass = 46 g/mol:
  \frac{2.3 \text{ g}}{46 \text{ g/mol}} = 0.05 \text{ mol} \; (\text{produces 0.05 mol of CO})
• For oxalic acid (H₂C₂O₄) with molar mass = 90 g/mol:
  \frac{4.5 \text{ g}}{90 \text{ g/mol}} = 0.05 \text{ mol} \; (\text{produces 0.05 mol of CO and 0.05 mol of CO}_2)

Total CO after CO₂ absorption by KOH = 0.05 \text{ mol (from formic acid)} + 0.05 \text{ mol (from oxalic acid)} = 0.10 \text{ mol} .

Finally, we calculate the weight of CO remaining:

The molar mass of CO is 28 g/mol. Therefore, the weight of 0.10 mol CO is:
\text{Weight} = 0.10 \text{ mol} \times 28 \text{ g/mol} = 2.8 \text{ g}

Thus, the weight of the remaining product at STP is 2.8 g. Therefore, the correct answer is option 2.8.