Question

1.5g of anhydrous cupric salt of an organic acid yields on thermal decomposition 0.392g of CuO calculate the equivalent weight of the acid (cu=63.6)

Answer 1

plzz help someone

Answer 2

Cu(X)2 + O2 --> CuO + CO2 + H2O
so n(CuO) = n(Cu(X)2)
n(CuO)=4,92*10^-3 mol
M(Cu(X)2) = m(Cu(X)2)/n(CuO)=304,8 gmol-1
M(X2)= M(Cu(X)2)-Ar(Cu) = 241,28 gmol-1
M(X)=120,64 gmol-1
if i got it right that is it...

Answer 3

the ans=121.5