Andrew C.

# How does these chemical equation balance, step by step? Why do they balance this way?

I already know how they balnace I just don't understand how they balance.

why does the O(2*) balance to 25? 2 C(8*)H(18*) + 25 0(2) yields 16 CO(2*) + 18 H(2*)O

These ones have me stumped as well

1 C(3)H(8) + 5 O(2) = 3 CO(2) + 4 H(2)O

Why does this equation balance this way?

1 Pb(OH)(2) + 2 HCl = 2 H(2)O + 1 PbCl(2)

Pb(OH)(2)  + HCl  + H(2)O + PbCl(2)

As I look at this equation, I am immediatly stumped because of the (OH). It kind of looks like it affects the whole equation, but I don't know why or know.

Please explain to me, in depth, why and how these equations balance this way

## 3 Answers By Expert Tutors

By: Tutor
5 (2)

Andrew C.

Thanks

In the equation below that one. How does that 5 get there before the O?

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11/17/12 David T.

tutor

For combustion reactions involving a simple hydrocarbon (like the first two equations you posted), you want to start by balancing the carbon, then the hydrogen, and finally the oxygen.

In the equation: C(3)H(8) + O(2) = CO(2) + H(2)O, you want to start by counting 3 C's on the reactant (left) side.  This means you need to add a coefficient of 3 to the product (right) side to make both sides equal.  So far it should look like this: 1 C(3)H(8) + O(2) = 3 CO(2) + H(2)O.  Next, we notice the 8 H's in the reactants, and add the coefficient of 4 to the water in the product side.  Now it looks like this: 1 C(3)H(8) + O(2) = 3 CO(2) + 4 H(2)O.  Now, count the O's on the product side: (3 x 2) + (4 x 1) = 10.  Since the reactant side has Oxygen as a diatomic molecule (meaning there is a subscript of 2 and therefore 2 times whatever coefficient we add), we need to place the coefficient 5 in front.  This gives us (5 x 2) = 10 O's on the product side.  Now simply go back and double check your quantities for C, H, and O individually to make sure it is balanced correctly, and you should be good to go.  Hope that helped clear things up a bit for you.  Good luck!

-David

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07/27/13 Tutor
4.9 (212)

Tutoring in General Chemistry, pre-Algebra, Algebra I and II, Reading Freddie P.

A correction to the atom balance in the first equation. 1C(3)H(8) + 5 O(2) = 3CO(2) + 4H(2)O C=1x3=3 C=3x1=3 H=1x8=8 H=4x2=8 O=5x2=10 O=(3x2)+(4x1)=10
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11/19/12 Tutor
5.0 (44)