a solution of 0.010M K2Cr2O7 with a volume of 60.00ml reacts in a titration reaction with 20.00ml of FeSO4 in sulphuric acid. Calculate the normality of FeSO4.

Answer 1

2 K2Cr2O7 + 3 FeSO4 + 14 H2SO4 = 2 K2SO4 + 2 Cr2(SO4)3 + 3 Fe(SO4)3 + 14 H2O

for every 2 mols of K2Cr2O7 you need 3 mols of FeSO4

You have .01*60*10^-3 mols of K2Cr2O7
so you need
.01(3/2)*60*10^-3 mols of Fe(SO4)3
in 20*10^-3 liters

.01(1.5)60/20 = .045 mols/liter

Answer 2

6 FeSO4 + K2Cr2O7 + 7 H2SO4 → K2SO4 + 3 Fe2[SO4]3 + Cr2[SO4]3 + 7 H2O
shouldn't this be the balanced chemical equation?
making it for every mole of K2Cr2O7 you need 6 moles of FeSO4??? and calculations can proceed from this information