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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? plz solv this que in easy formula with full solution.?
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A mixture of 2.3 g formic acid and 4.5 g oxalic acid is treated with c...
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A mixture of 2.3 g formic acid and 4.5 g oxalic acid is treated with c...
Given:
Mass of formic acid (HCOOH) = 2.3 g
Mass of oxalic acid (H2C2O4) = 4.5 g

To find:
Weight of the remaining product at STP after passing the evolved gaseous mixture through KOH pellets.

Solution:
1. Chemical Reaction:
The chemical reaction involved in this process is the decomposition of formic acid and oxalic acid by concentrated sulfuric acid (H2SO4):

HCOOH + H2SO4 → CO + H2O + SO2
H2C2O4 + H2SO4 → CO2 + H2O + SO2

2. Calculating Moles:
First, we need to calculate the number of moles of formic acid and oxalic acid present in the given masses. We can use the formula:

Number of moles = Mass / Molar mass

The molar mass of formic acid (HCOOH) is 46.03 g/mol, and the molar mass of oxalic acid (H2C2O4) is 90.03 g/mol.

Number of moles of formic acid = 2.3 g / 46.03 g/mol
Number of moles of oxalic acid = 4.5 g / 90.03 g/mol

3. Calculating the Limiting Reactant:
To determine the limiting reactant, we compare the number of moles of each reactant. The reactant that produces the lesser amount of product is the limiting reactant.

Now, we need to calculate the number of moles of CO and CO2 produced from the complete reaction of formic acid and oxalic acid, respectively. The balanced equations tell us that one mole of formic acid produces one mole of CO, and one mole of oxalic acid produces one mole of CO2.

Number of moles of CO = Number of moles of formic acid
Number of moles of CO2 = Number of moles of oxalic acid

4. Calculating Weight of Evolved Gases:
Next, we need to calculate the weight of the evolved gases (CO and CO2) produced from the complete reaction of formic acid and oxalic acid.

Weight of CO = Number of moles of CO × Molar mass of CO
Weight of CO2 = Number of moles of CO2 × Molar mass of CO2

The molar mass of CO is 28.01 g/mol, and the molar mass of CO2 is 44.01 g/mol.

5. Calculating Weight of Remaining Product:
Finally, we need to calculate the weight of the remaining product after passing the evolved gaseous mixture through KOH pellets. The evolved gases (CO and CO2) react with KOH to form potassium formate (HCOOK) and potassium oxalate (K2C2O4), respectively.

The balanced equations for these reactions are:

CO + 2KOH → HCOOK + K2O
CO2 + 2KOH → K2C2O4 + H2O

The weights of the remaining products can be calculated using the number of moles of CO and CO
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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? plz solv this que in easy formula with full solution.?
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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? plz solv this que in easy formula with full solution.? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about 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? plz solv this que in easy formula with full solution.? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 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? plz solv this que in easy formula with full solution.?.
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