A mixture of 2.3 g formic acid and 4.5 oxalic acid is treated with con

1.	A mixture of 2.3 g formic acid and 4.5 oxalic acid is treated with conc.H2SO4 . The evolved gaseous mixture is passed through KOH pallets . Weight in (g) of the remaining product at STP will be
Answer» Answer: 2.8g Explanation: HCOOH → CO+H2O (MOLES)i = 2.346 = 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 Thus, the remaining product will be only CO Moles of CO formed from both REACTIONS - = 1/20 + 1/20 = 110 Left mass of CO = moles × MOLAR mass = 1/10×28 = 2.8g Thus, the remaining weight will be 2.8g

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

Answer»

Answer:

2.8g

Explanation:

HCOOH → CO+H2O

(MOLES)i = 2.346 = 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

Thus, the remaining product will be only CO

Moles of CO formed from both REACTIONS -

= 1/20 + 1/20 = 110

Left mass of CO = moles × MOLAR mass

= 1/10×28

= 2.8g

Thus, the remaining weight will be 2.8g