Question

a mixture of 2.3 g formic acid and 4.5 g oxalic acid is treated with conc. h2so4. the evolved gaseous mixtue is passed through koh pellets .weight of the remaining product at stp will be​

Answer 1

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

Answer 2

Answer: 2.8 g

Explanation:check out the attachment given below