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1 Jawaban
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1. Jawaban chemicalromance
oksidator :
Fe^2+ > Fe^3+ + e^-
reduktor :
Cr2O7^2- + 14H^+ + 8e^- > 2Cr^2+ + 7H2O
8Fe^2+ > 8Fe^3+ + 8e^- |×8|
Cr2O7^2- + 14H^+ + 8e > 2Cr^2+ + 7H2O |×1|
_______________________________
8Fe^2+ + Cr2O7^2- + 14H^+ > 8Fe^3+ + 2Cr^2+ + 7H2O
mol FeSO4 = 0,03 x 0,2 = 0,006mol
mol Cr2O7^2- = 1/8 x 0,006 = 0,00075mol
volume Cr2O7^2- = n/M
volume Cr2O7^2- = 0,00075/0,1 = 0,0075L = 7,5mL