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A mixture of 2.3 g formic acid and 4.5 g...

A mixture of 2.3 g formic acid and 4.5 g oxalic acid is treated with conc. `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

Text Solution

Verified by Experts

The correct Answer is:
C
2.3 gm formic acid (HCOOH)
` = (2.3)/(46) = 1/20 ` mol HCOOH
4.5 gram oxalic acid `(H_2 C_2O_4)
` = (4.5)/(90) = 1/20 ` mol `H_2C_2O_4`
(i) `underset(1/20 "mol")(HCOOH) overset(conc. H_2SO_4)(to) underset(1/20 "mol")(CO + H_2O)`
(ii) `H_2C_2 O_4 overset(conc. H_2SO_4)(to) CO + CO_2 + H_2O`
`CO_2` will be absorbed in KOH anc CO is additional product.
Total moles of additional `CO_2 = 1/20 + 1/20 = 1/10` mol
` = 1/10 "mol" = 1/10 xx 28 = 2.8 g`
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Knowledge Check

  • Which is the conjugate acid of H_2SO_4 ?

    A
    `HSO_4^-`
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    `H_3O^+`
    C
    `H_3SO_4^+`
    D
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  • A gaseous mixture consists of 16 g of helium (He) and 16 g oxygen (O_2) , the ratio C_P/C_V of the mixture is = ......

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    1.4
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    1.54
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