Asked by malissa
A 3.75-G SAMPLE OF IRON ORE IS TRANSFORMED TO A SOLUTION OF IRON(II)SULFATE, FeSO4,AND THIS SOLUTION IS TITRATED WITH 0.150 M K2Cr2O7. IF IT REQUIRES 43.7 mL OF POTASSIUM DICHROMATE SOLUTION TO TITRATE THE IRON(II)SULFATE SOLUTION, WHAT IS THE PERCENTAGE OF IRON IN THE ORE? THE REACTION IS
6FeSO4(aq)+K2Cr207(aq)+7H2SO4(aq)¨
3Fe2(SO4)3(aq)+Cr2(SO4)3(aq)+7H2O(l)+K2SO4(aq)
6FeSO4(aq)+K2Cr207(aq)+7H2SO4(aq)¨
3Fe2(SO4)3(aq)+Cr2(SO4)3(aq)+7H2O(l)+K2SO4(aq)
Answers
Answered by
DrBob222How
mols dichromate = M x L = ?
Use the coefficients in the balanced equation to convert mols dichromate to mols FeSO4
Convert mols FeSO4 to mols Fe then to grams Fe. g Fe = mols Fe x atomic mass Fe.
%Fe = (mass Fe/mass sample)*100 = ?
Use the coefficients in the balanced equation to convert mols dichromate to mols FeSO4
Convert mols FeSO4 to mols Fe then to grams Fe. g Fe = mols Fe x atomic mass Fe.
%Fe = (mass Fe/mass sample)*100 = ?
Answered by
Geovanna
62.81%
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.