Hi I have this pre-lab question that needs to be quickly answered. Please help..
I need to find the coefficients for this equation, I've tried couple of times but didn't get the answer.
K2Cr2O7 + H2SO4 + C2H5OH ---> Cr2(SO4)3 + K2SO4 + CH3COOH + H2O
Thank you
2007-11-24 10:52:48 · 2 answers · asked by miss_frd 1 in Science & Mathematics ➔ Chemistry
Use half-equations
Cr2O72- + 14H+ + 6e- --> 2Cr3+ + 7H2O
C2H5OH + H2O --> CH3COOH + 4H+ + 4e-
Multiply the first by 2 and the second by 3, then add:
2Cr2O72- + 16H+ + 3C2H5OH --> 4Cr3+ + 11H2O + 3CH3COOH
Add in the K+ and SO42-
2 K2Cr2O7 + 8H2SO4 + 3C2H5OH --> 2Cr2(SO4)3 + 2K2SO4 + 11H2O + 3CH3COOH
2007-11-24 11:18:23 · answer #1 · answered by Chemmunicator 5 · 0⤊ 0⤋
I think this is the balanced equation:
K2Cr2O7 + 4 H2SO4 + 3 C2H5OH --->
Cr2(SO4)3 + K2SO4 + 3 C2H3OOH + 4 H2O
Added: I think the other guy has got it; I must have had a total "brain-fart" on this one! I wonder who I should contact about having a medal taken away?
2007-11-24 19:09:44 · answer #2 · answered by Flying Dragon 7 · 0⤊ 1⤋