Anyone know how to balance equations

may you please help me with balancing equations you have to use the lowest possible coefficients and Include states-of-matter under the given conditions in your answer.
C2H6(g) + O2(g) → CO2(g) + H2O(g)
CO2(g) + H2O(l) → C6H12O6(s) + O2(g)
thanks!:)

steps to balancing a non-redox rxn..
(1) write down the equation with blanks where the coefficients will eventually end up.
(2) choose a starting element. pick on that exists in 1 reactant and 1 product if possible. Avoid H and O if possible
(3) choose a starting number. Any will do. The usual ones are 1,2,3, or 4. Your choice of starting numbers could result in you working in fractions and having to eliminate them at the end.
(4) put your starting number in the blank in front of the reactant containing your starting element
(5) balance your starting element on the right.
(6) go back and forth balancing counter ions. ie other elements in the same reactant or product. C2H6.. if I balance C, the counter ion is H

************
__ C2H6(g) + __ O2(g) → __ CO2(g) + __ H2O(g)

I'll choose "C" and the number 1
1 C2H6(g) + __ O2(g) → __ CO2(g) + __ H2O(g)

1x2 = 2 C's on the left.. I need 2 on the right
1 C2H6(g) + __ O2(g) → 2 CO2(g) + __ H2O(g)

1x6 = 6 H's on the left.. I need 6 on the right
1 C2H6(g) + __ O2(g) → 2 CO2(g) + 3 H2O(g)

2x2 + 3x1 = 7 H's on the RIGHT.. So I need 7 on the left
1 C2H6(g) + 7/2 O2(g) → 2 CO2(g) + 3 H2O(g)

multiply through by 2 to get rid of the fraction
2 C2H6(g) + 7 O2(g) → 4 CO2(g) + 6 H2O(g)

******************
__ CO2(g) + __ H2O(l) → __ C6H12O6(s) + __ O2(g)

I'll start with C and the number "6".. why? if the coefficient of C6H12O6 = 1, then I'll need 6 CO2's on the left.. make sense?
6 CO2(g) + __ H2O(l) → __ C6H12O6(s) + __ O2(g)

6 C's on the left.. need 6 on the right... and 1 C6H12O6 has 6 C's in it!
6 CO2(g) + __ H2O(l) → 1 C6H12O6(s) + __ O2(g)

1x12 = 12 H's on the right.. so I need 12 on the left.
6 CO2(g) + 6 H2O(l) → 1 C6H12O6(s) + __ O2(g)

and finally 6x2 + 6x1 = 18 O's on the left.. I need 18 on the right. I already have 1x6.. so I need 12 more
6 CO2(g) + 6 H2O(l) → 1 C6H12O6(s) + 6 O2(g)

2C2H6 + 5O2 --> 4CO2 + 6H2O
make the atoms on the right = the atoms on the left

6CO2 + 6H2O --> C6H12O6 + 6O2

carbon:

C2H6(g) + O2(g) → 2CO2(g) + H2O(g)

hydrogen:

C2H6(g) + O2(g) → 2CO2(g) + 3H2O(g)

oxygen:

C2H6(g) + 7/2 O2(g) → 2CO2(g) + 3H2O(g)

clear fraction:

2C2H6(g) + 7O2(g) → 4CO2(g) + 6H2O(g)