Question

0.9 g oxalic acid is treated with conc h2so4 the evolved gas is passed through naoh pellets . what is remaining gaseous product in stp?

Answer 1

when oxalic acid is treated with conc. H2SO4 , the dissociation of oxalic acid will be .....

H2C2O4 ⇔H2O + CO2(↑) + CO(↑)

now evolved gas {i.e., (CO2 + CO) } passing through NaOH pallets, then CO2 gas is absorbed and remaining CO2 in the container.

now no of mole of oxalic acid = given weight/molecular weight of oxalic acid

= 0.9/90 = 0.01 mol

from reaction, one mole of oxalic acid produces one mole of CO.

so, no of mole of CO = no of mole of oxalic acid = 0.01 mol

now, at STP volume of remaining gaseous product = volume of CO = 0.01 × 22.4L

= 0.224 L

= 224 ml