Question

A mixture of $2.3 \,g$ formic acid and $4.5 \,g$ oxalic acid is treated with cone. $H_2SO_4$. The evolved gaseous mixture is passed through $KOH$ pellets. Weight (in $g$ ) of the remaining product at $STP$ will be

• A. 4.4
• B. 1.4
• C. 2.8
• D. 3

Answer 1

The Correct Option is C

To solve this problem, we need to understand the chemical reaction that takes place, calculate the number of moles of gases evolved, and finally determine the weight of the remaining product after the gases have reacted with KOH.

Given:

• Formic acid mass: 2.3 \, \text{g}
• Oxalic acid mass: 4.5 \, \text{g}

The reaction of formic acid and oxalic acid with concentrated sulphuric acid evolves carbon monoxide (\text{CO}) and carbon dioxide (\text{CO}_2) gases. The respective reactions are as follows:

1. Formic acid:
   \text{HCOOH} \rightarrow \text{CO} + \text{H}_2\text{O}
2. Oxalic acid:
   \text{C}_2\text{H}_2\text{O}_4 \rightarrow \text{CO}_2 + \text{CO} + \text{H}_2\text{O}

Next, we calculate the moles of formic and oxalic acid:

1. Molar mass of formic acid (\text{HCOOH}): 46 \, \text{g/mol}
2. Molar mass of oxalic acid (\text{C}_2\text{H}_2\text{O}_4): 90 \, \text{g/mol}

Calculate moles:

• Moles of formic acid: \frac{2.3}{46} = 0.05 \, \text{mol}
• Moles of oxalic acid: \frac{4.5}{90} = 0.05 \, \text{mol}

The evolved gas from formic acid is 0.05 \, \text{mol} \, \text{CO}. From oxalic acid, we have 0.05 \, \text{mol} \, \text{CO} and 0.05 \, \text{mol} \, \text{CO}_2. Thus, total moles of gas:

0.05 + 0.05 + 0.05 = 0.15 \, \text{mol}

Upon passing the gaseous mixture through KOH, only \text{CO}_2 is absorbed (since KOH absorbs \text{CO}_2, not \text{CO}). Therefore, 0.05 \, \text{mol} \, \text{CO}_2 is absorbed, leaving behind 0.1 \, \text{mol} \, \text{CO}.

Calculate mass of remaining \text{CO} gas at STP:

Molar mass of \text{CO}: 28 \, \text{g/mol}

Mass of \text{CO}: 0.1 \times 28 = 2.8 \, \text{g}

Therefore, the weight of the remaining product at STP is 2.8 g.