Best Answer:
to balance ordinary chemical reactions follow these steps

(1) write down all the reactants and products leaving blanks before them

(2) choose a starting element. pick one that is in 1 reactant and 1 product if possible. avoid H and O

(3) choose a starting number... 1,2,3,4 are all typical and good starting #'s

(4) put your choice number in front of the REACTANT containing your element of choice

(5) balance the # of atoms of that element on the product side

(6) go back and forth balancing counter ions... (other elements in the reactant or product you're working with)

here we go...

__ NH2CH2COOH (s) + __ O2 -----> __ CO2 (s) + __ H2O (l) + __ N2 (g)

so I'll pick N and the #1 to start with...

1 NH2CH2COOH (s) + __ O2 -----> __ CO2 (s) + __ H2O (l) + __ N2 (g)

then balancing N.. 1N on the left... and oops.. I have N2 on the right.. so 1/2 x N2 = 1N right?

1 NH2CH2COOH (s) + __ O2 -----> __ CO2 (s) + __ H2O (l) + 1/2 N2 (g)

balancing H's.. 1x2 +1x2 + 1x1 = 5 on the left.. so I need 5 on the right.. 5/2 x H2O gives me 5 H's on the right.. right?

1 NH2CH2COOH (s) + __ O2 -----> __ CO2 (s) + 5/2 H2O (l) + 1/2 N2 (g)

next... I'll balance C's.. 2 on the left.. so I need 2 on the right

1 NH2CH2COOH (s) + __ O2 -----> 2 CO2 (s) + 5/2 H2O (l) + 1/2 N2 (g)

and that just leaves O.. 2x2 + 5/2x1 = 4 + 5/2 = 8/2 + 5/2 = 13/2's on the right.. so I need 13/2's on the left.. BUT my 1 NH3CH3CO2H already has 2 O's.. so I need 13/2 - 2 = 13/2 - 4/2 = 9/2's more on the left....9/4 x O2 gives me 9/2 O's

and tada

1 NH2CH2COOH (s) + 9/4 O2 -----> 2 CO2 (s) + 5/2 H2O (l) + 1/2 N2 (g)

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now we could have avoided the fractions by picking a different starting number. If you look at the N2 on the right.. it's pretty clear that if I pick N and #1, I'm going to get a fraction!.. so maybe I should have seen that and started with the # "2"..

so I'll pick N and the #1 to start with...

2 NH2CH2COOH (s) + __ O2 -----> __ CO2 (s) + __ H2O (l) + __ N2 (g)

2 NH2CH2COOH (s) + __ O2 -----> __ CO2 (s) + __ H2O (l) + 1 N2 (g) .. balancing N

2 NH2CH2COOH (s) + __ O2 -----> __ CO2 (s) + 5 H2O (l) + 1 N2 (g).. balancing H

2 NH2CH2COOH (s) + __ O2 -----> 4 CO2 (s) + 5 H2O (l) + 1 N2 (g) .. balancing C

2 NH2CH2COOH (s) + 9/2 O2 -----> 4 CO2 (s) + 5 H2O (l) + 1 N2 (g) .. balancing O

And that's probably the best starting guess you could make in which case you then go on and double the equation to get

4 NH2CH2COOH (s) + 9 O2 -----> 8 CO2 (s) + 10 H2O (l) + 2 N2 (g) .. balancing O

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here's the points you should take away

(1).. follow the steps I outlined above

(2).. the choice of starting number makes the problem easier or harder..

(3).. fractions make balancing difficult, but if done correctly the RATIOS will be correct and that's all you really care about!

(4).. usually, and you can see that with the other answerers, you want whole number coefficients in the balanced equation

and of course.. those coefficients are mole ratios!

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