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_(2)SO_(4)`. The evolved gaseous mixture is passed through KOH pellets. Weight (in g) of the remaining product at STP will be

Text Solution

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The correct Answer is:

2.8

`underset(underset(46 g)("Formic acid"))(HCOOH) overset(conc.H_(2)SO_(4))rarrunderset(28g)(CO)+ H_(2)O`
Now 46 g of COOH evolve CO = 28 g
`:. 2.3 g` of HCOOH will evolve `CO = (28)/(46) xx 2.3`
= 1.4 g
`{:(""COOH),("| " overset(conc.H_(2)SO_(4))rarr CO+CO_(2)+H_(2)O),(COOH" "28 g),("Oxalic acid"),(" "90g):}`
When the gaseous mixture of `(CO+CO_(2)` is passed through KOH pellets, `CO_(2)` is absorbed while CO phase out
`2KOH + CO_(2) rarr K_(2)CO_(3) + H_(2)O`
Now, 90g of oxalic acid evolve `CO = 28 g`
`:. 4.5 g` of oxalic acid will evolve CO
`= (28)/(90) xx 4.5 = 1.4 g`
Total amount of CO evolved = 1.4 + 1.4 = 2.8 g

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When formic acid or oxalic acid is treated with conc. H_(2)SO_(4) , the gas evolved is

`H_(2)S`

`SO_(2)`

`CO_(2)`

Oxalic acid on treatment with conc. H_(2) SO_(4) gives

CO only

`CO_(2)` only

`CO_(2) + H_(2)O`

`H_(2)O + CO + CO_(2)`

When formic acid is treated with conc. H_(2)SO_(4) , the gas evolved is :