I am having a really hard time trying to balance these 3 formulas. Please help me!!?

Question

1. CH3 OH + O2 --> CO2 + H2 O

2. Al + HNO3 --> H3 + Al(NO3)3

3. (NH4)2 Cr2 07 --> Cr2 O3 + N2 + H2O

All of the numbers go below the previous letter. Like H(2 below) and O

Answer 1

For the first one; when you try to balance the equation with just one CH3OH you find that you need1 & 1/2 an O2 get the dang thing to balance, when this happens, to fix it, double the number of CH3OH's and you should get to an even number of oxygen molecules.

CH3OH + 1 & 1/2 O2 ---> CO2 + 2 H2O (no good)

2 CH3OH +3 O2 --->2 CO2 + 4 H2O (OK)

For #2 you have H3 I presume you mean H2;

Again, here you wind up having to double it again but this time it's because one of the products is comming out a fraction

Al + 3 HNO3 ---> 1 & 1/2 H2 + Al(NO3)3 (no good)

2Al +6 HNO3 --->3 H2 + 2 Al(NO3)3 (OK)

For the last one:

This one is pretty easy; you pull out the Cr2O3 from the starting material formula

(NH4)2Cr2O7 ---> Cr2O3 + "(NH4)2O4"

Now, pull out the N2:

(NH4)2Cr2O7 ---> Cr2O3 + N2 + "(H4)2O4"

Then figure out how many waters can be formed from "(H4)2O4" 8 H's + 4 O's (Since it comes out even [no extra atoms left over] you don't need to double anything).

(NH4)2Cr2O7 ---> Cr2O3 + N2 +4 H2O

Answer 2

1. CH3OH + 3/2 O2 -> CO2 + 2H2O
2. Al + 3 HNO3 -> H3 + Al(NO3)3
3. (NH4)2Cr2O7 -> Cr2 + N2 + 4 H20