A mixture of 2.3 gram formic acid and 4.5 gram oxalic acid is treated with concentrated h2 s o4 the evolved gas mixture is produced through pellets weight in gram of the remaining product at stp will be
Answers
Answer:
2.8 grams is the correct answer.
Explanation:
Given data:
Weight of formic acid = 2.3 grams
Weight of oxalic acid = 4.5 grams
It is treated with concentrated H2SO4.
THEe evolved gas mixture is produced through pellets.
To find :
weight in gram of the remaining product at stp =?
Solution:
The reaction involved in this process is as given below
HCOOH ===> CO + H2O
Initially the moles are zero.
In final conditions, the moles of CO are 1/20 and that of H2O are 1/20.
The next reaction in the process involved will be
H2C2O4 ===> CO2+ CO + H2O
The initial number of moles are zero.
In the final condition the moles of CO2, CO and H2O are 1/20 each.
Now in this process, CO only remains.
This is because CO2 is absorbed by KOH.
The total numbers if moles formed will be
= 0 + 1/20 +1/20
= 2/20
= 1/10
= 0.1
Now we have to calculate the leftass of CO2
It is given by the formula
Left mass
= molecular mass * Number of moles
= 28 * 0.1
= 2.8 grams of mass is left.
Hence,
2.8 grams is the correct answer.
Given :
Weight of formic acid = 2.3 g
Weight of Oxalic acid = 4.5 g
Solution:
HCOOH ⟶ CO + H2O
(moles)i = 2.3 / 46 = 1 / 20 0 0
(moles)i= 0 1 / 20 1 /20
H₂C₂O₄ H₂SO₄------→CO + CO₂ + H₂O
(moles)i=4.5 / 90 = 1 / 20 0 0 0
(moles)i=0 1 / 20 1 / 20 1 / 20
CO₂ is absorbed by KOH
So the remaining product is only CO
Moles of CO formed from both reactions.
=1 / 20 + 1 / 20 = 1 / 10
Left mass of CO=moles×molarmass
=1 / 10×28
Left mass of CO=2.8 g