A mixture of 2.3g formic acid and 4.5g oxalic acid (anhyd.) is treated with conc. H2SO4 . The evolved gaseous mixture is passed through KOH pellets. Weight (in g) of the remaining gaseous product at STP will be

Question

A mixture of 2.3g formic acid and 4.5g oxalic acid (anhyd.) is treated with conc. H2SO4 . The evolved gaseous mixture is passed through KOH pellets. Weight (in g) of the remaining gaseous product at STP will be (A) 2⋅ 8 (B) 3⋅ 0 (C) 1⋅ 4 (D) 4⋅ 4

Tardigrade Admin · Accepted Answer

HCOOH arrow H2O(.l.)+CO(.g.) 2.3gm((1/20) mole) H2C2O4 arrow H2O(.l.)+CO(.g.)+(CO)2(.g.) (1/20) text mole ((1/20) + (1/20))moleCO,(1/20)mole(CO)2 CO2 is absorbed by KOH So, (1/10)mole of CO is left =2.8gm of CO

Q. A mixture of $2.3 g$ formic acid and $4.5 g$ oxalic acid (anhyd.) is treated with conc. $H_{2} S O_{4}$ . The evolved gaseous mixture is passed through $K O H$ pellets. Weight (in g) of the remaining gaseous product at $STP$ will be

$2 \cdot 8$

$3 \cdot 0$

$1 \cdot 4$

$4 \cdot 4$

Q. A mixture of 2.3g formic acid and 4.5g oxalic acid (anhyd.) is treated with conc. H2​SO4​ . The evolved gaseous mixture is passed through KOH pellets. Weight (in g) of the remaining gaseous product at STP will be

2⋅8

3⋅0

1⋅4

4⋅4

HCOOH→H2​O(l)+CO(g) 2.3gm(201​mole) H2​C2​O4​→H2​O(l)+CO(g)+(CO)2​(g) 201​ mole (201​+201​)moleCO,201​mole(CO)2​ CO2​ is absorbed by KOH So, 101​mole of CO is left =2.8gm of CO

Q. A mixture of $2.3 g$ formic acid and $4.5 g$ oxalic acid (anhyd.) is treated with conc. $H_{2} S O_{4}$ . The evolved gaseous mixture is passed through $K O H$ pellets. Weight (in g) of the remaining gaseous product at $STP$ will be

$2 \cdot 8$

$3 \cdot 0$

$1 \cdot 4$

$4 \cdot 4$