Answer:
1.4 × 10^-4 M
Explanation:
The balanced redox reaction equation is shown below;
5Fe2+ + MnO4- + 8H+ --> 5Fe3+ +Mn2+ + 4H2O
Molar mass of FeSO4(NH4)2SO4*6H2O = 392 g/mol
Number of moles Fe^2+ in FeSO4(NH4)2SO4*6H2O = 3.47g/392g/mol = 8.85 × 10^-5 moles
Concentration of Fe^2+ = 8.85 × 10^-5 moles × 1000/200 = 4.425 × 10^-4 M
Let CA be concentration of Fe^2+ = 4.425 × 10^-4 M
Volume of Fe^2+ (VA)= 20.0 ml
Let the concentration of MnO4^- be CB (the unknown)
Volume of the MnO4^- (VB) = 12.6 ml
Let the number of moles of Fe^2+ be NA= 5 moles
Let the number of moles of MnO4^- be NB = 1 mole
From;
CAVA/CBVB = NA/NB
CAVANB = CBVBNA
CB= CAVANB/VBNA
CB= 4.425 × 10^-4 × 20 × 1/12.6 × 5
CB = 1.4 × 10^-4 M