Balance the following equation in acidic conditions.
cr2+ + h2moo4 --> cr3+ mo

Cr^2+ + H2MoO4 + H+ → Cr^3+ + Mo + H2O 
Cr goes from +2 to +3 or loses one e- 
Mo is +6 in H2MoO4 so it goes to 0 gaining 6 e- 
so we must multiply Cr by 6 to balance e- 

6Cr^2+ + H2MoO4 + H+ → 6Cr^3+ + Mo + H2O 
now we balance the spectators 
take O first - 4 on left then, change the right side 

6Cr^2+ + H2MoO4 + H+ → 6Cr^3+ + Mo + 4H2O 
now we have 8 H's on the right 
we don't want to touch the H2MoO4 
so we make H+ up to 6 for a total of 8 on the left 

6Cr^2+ + H2MoO4 + 6H+ → 6Cr^3+ + Mo + 4H2O 
Then, It's balanced

Ci think not