How many grams of iron(2) sulfate heptahydrate (pH<7) can be oxidized with 28 cm3 potassium permanganate solution ,concentration of 0,02 mol/dm3.
I am aware of that this problem is actually so easy but there is no way that my result matches with the result in the book
I wrote the equation,like this:
2 KMnO4 + 10 FeSO4 + 8 H2SO4 --> 5 Fe2(SO4)3 + 2 MnSO4 + 1 K2SO4 + 8 H2O
H2SO4 is needed because acidic environment is needed for the reaction to take place. There is no way that iron (2) sulfate-heptahydrate can directly oxidize with KMnO4.I am sure that this equation is correct. Than I founded the ammount of KMnO4.
c=n/V
n=c*V=0,02*0,028=0,00056 mol
But I can not determine the mole ratio,according to the chemical reaction, I am so confused
n(FeSO4*7H2O); n(KMnO4)=x:x
The result should be: 0,7747g of FeSO4*7H2O