Question

1. (a) What are the decomposition products when the copper oxalate hydrate is heated in the test tube for this experiment ?

(b) What is the balanced chemical reaction between oxalate and permanganate in the titration reaction ?

(c) What is the mole ratio between oxalate & permanganate ? ................................C2O₄ ^-2 : ...................................MnO₄ ^-1

(d) How is the percent water in copper oxalate hydrate determined in this experiment?

Answer 1

Part a

Decomposition of copper oxalate hydrate gives copper oxalate and water

CuC2O4.xH2O = CuC2O4 + xH2O

Part b

The balanced chemical reaction

5C2O4(-2) + 2MnO4(-) + 16H+ = 10CO2 + 2Mn(2+) + 8H2O

Part c

Mol ratio of C2O4(-2) : MnO4(-) = 5:2 = 2.5 : 1

Part d

% water in hydrate

= (mass of water removed on heating x 100) / (mass of copper oxalate hydrate)

For example

For copper oxalate Mono hydrate

CuC2O4.H2O = CuC2O4 + H2O

% water = 18*100/(18+151.56)

= 10.62%