Chemistry, asked by adam694, 1 year ago

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

Answered by Anonymous
6

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.

Answered by bestanswers
6

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

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