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A mixture of 2.3 2.3 2.3 g formic acid and 4.5 4.5 4.5 g oxalic acid is treated with conc. H 2 S O 4 . H_2SO_4. H 2 S O 4 . The evolved gaseous mixture is passed through KOH pellets. Weight (in g) of the remaining product at STP will be
hard
Basic Concepts
2018
chemistry
Explanation To solve this problem, we need to understand the chemical reactions involved when formic acid and oxalic acid are treated with conc. H 2 S O 4 . H_2SO_4. \\
H 2 S O 4 . 1. Reaction of formic acid with conc. H 2 S O 4 : H_2SO_4: \\
H 2 S O 4 :
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Formic acid
decomposes to produce carbon monoxide
and water.
H C O O H → C O + H 2 O \\
HCOOH \rightarrow CO + H_2O \\
H COO H → CO + H 2 O 2. Reaction of oxalic acid with conc.
H 2 S O 4 : H_2SO_4: \\
H 2 S O 4 : Oxalic acid
( H 2 C 2 O 4 ) (H_2C_2O_4) ( H 2 C 2 O 4 ) decomposes to produce carbon dioxide
carbon monoxide
and water.
H 2 C 2 O 4 → C O + C O 2 + H 2 O \\
H_2C_2O_4 \rightarrow CO + CO_2 + H_2O \\
H 2 C 2 O 4 → CO + C O 2 + H 2 O 3. The gaseous mixture evolved consists of
and
When this mixture is passed through KOH pellets,
is absorbed by KOH, leaving only
Let's calculate the moles of each acid:
• Molar mass of formic acid
( H C O O H ) = 46 (HCOOH) = 46 \, ( H COO H ) = 46 g/mol
• Moles of formic acid
= 2.3 46 = 0.05 = \frac{2.3}{46} = 0.05 \, = 46 2.3 = 0.05 mol
• Molar mass of oxalic acid
( H 2 C 2 O 4 ) = 90 (H_2C_2O_4) = 90 \, ( H 2 C 2 O 4 ) = 90 g/mol
• Moles of oxalic acid
= 4.5 90 = 0.05 = \frac{4.5}{90} = 0.05 \, = 90 4.5 = 0.05 mol
4. Calculate the moles of
produced:
• From formic acid:
Each mole of formic acid produces 1 mole of
Moles of
from formic acid
mol
• From oxalic acid:
Each mole of oxalic acid produces 1 mole of
Moles of
from oxalic acid
mol
• Total moles of
C O = 0.05 + 0.05 = 0.10 CO = 0.05 + 0.05 = 0.10 \, CO = 0.05 + 0.05 = 0.10 mol
5. Calculate the weight of
at STP:
• Molar mass of
g/mol
• Weight of
C O = 0.10 × 28 = 2.8 CO = 0.10 \times 28 = 2.8 \, CO = 0.10 × 28 = 2.8 g
Therefore, the weight of the remaining product (CO) at STP is
g
This corresponds to Option 1.