Asample of 1.55 g of iron ore is dissolved in an acid solution in which the iron is converted into
fe2+. the solution formed is then titrated with kmno4 which oxidises fe2+ to fe3+ while the mno4-
ions are reduced to mn2+ ions. 92.95 ml of 0.020 m kmno4 is required for the titration to reach
the equivalence point.
a) write the balanced equation for the titration.
b) calculate the percentage of iron in the sample.

a) 8H2SO4+2KMnO4+10FeSO4→5Fe2(SO4)3+8H2O+2MnSO4+K2SO4

B) nKMno4 = 0.001859 mol

=>nFeso4=0.009295 mol

nFe= 0.009295 mol

mFe=0.52052 g

=>percentage of iron in the sample: 33.5819%

Explanation:a) 8H2SO4+2KMnO4+10FeSO4→5Fe2(SO4)3+8H2O+2MnSO4+K2SO4

B) nKMno4 = 0.001859 mol

=>nFeso4=0.009295 mol

nFe= 0.009295 mol

mFe=0.52052 g

i just calculated it all out, and not including weekends, and all the different days off. as of this school year (2018-2019) students are spending 2198 hours in school this year