Balance the following reaction under acidic conditions. Cr2O7 2- + Mo → CrO2 - + MoO+

first we look at both sides....
then we take Cr...and note the number it carries (which is two)
so in order to balance that we check if the Cr in the other end of the reaction is 2 as well...and since it's not..we add the 2
then it'll give Cr2 --> 2Cr
(note you can't add numbers after the symbol it's always before it)
so now that is balanced then we look at O (which is the next) it has 7 after
then we check the O in the other end...(we see that it is O2) but the fact that it is joined with the Cr that you added 2 to...the two affects it and the O is also multiplied by 2 along with the Cr
so therefore the O in the other end of the reaction is equivalent to 4...and that will be
Cr2O7 2- --> 2CrO2
but the O is still not balanced...then we see that O appeared twice in the other end...
so we can balance the O in the product side using the two O's
then we think...what number will make this O = 7..when it is = 4...

and that is 3...

so we add 3 in front of the compound carrying the O
to make it 3MoO...
now the O is balanced.... but the Mo is not balanced.. and we have to balance it as well...
so we look at the reactant side and see that Mo= 1 while in the product side Mo= 3..
so the simple thing to do is add 3 IN FRONT of the Mo in the reactant side
and that will be 3Mo

so in all everything is balanced

and it's will be

I hope you understandᥬ᭄