# Homework Help: Help with balancing a chemical equation algebraically

1. Dec 27, 2012

### v3ra

1. The problem statement, all variables and given/known data

(a)H2SO4 + (b)NaHCO3 -> (x)Na2SO4 + (y)CO2 + (z)H2O

I gave each element it`s own equation:
(1) For hydrogen – 2a + b = 2z
(2) For sulphur – a = x
(3) For oxygen – 4a + 3b = 4x + 2y + z
(4) For sodium – b = 2x

I then assigned a number to letter 'a'. My book says to begin with the number 12 because it is the most conventionally used.

I began with equation (2) since it had the least amount of variables and proceeded in this fashion.

a = 12

(2)
a = x
x = 12

(4)
b = 2x
b = 2(12)
b = 24

(1)
2a + 2b = 2z
2(12) + 2(24) = 2z
24 + 48 = 2z
72 = 2z
z = 36

(3)
4a + 3b = 4x + 2y + z
4(12) + 3(24) = 4(12) + 2y + 36
48 + 72 = 48 + 2y + 36
120 = 84 + 2y
36 = 2y
y = 18

I then divided each of the values by a common divider, in this case '6' in order to get the lowest possible coefficient. I ended up with:
a = 2
b = 4
x = 2
y = 3
z = 6

When I add these to my original formula, it is not balanced:
(2)H2SO4 + (4)NaHCO3 -> (2)Na2SO4 + (3)CO2 + (6)H2O

I also tried dividing by 12, 4 and 2 and I am still left with an unbalanced formula.

The book says that when this happens, I must instead assign the value of 1 to 'a', I did this and got the following results:

a = 1

(2)
a = x
x = 1

(4)
b = 2x
b = 2(1)
b = 2

(1)
2a + 2b = 2z
2(1) + 2(2) = 2z
2 + 4 = 2z
6 = 2z
z = 3

(3)
4a + 3b = 4x + 2y + z
4(1) + 3(2) = 4(1) + 2y + 3
4 + 6 = 4 + 2y + 3
10 = 7 + 2y
3 = 2y
y = 1.5

As you can see, 'y' is left at 1.5 which is not a whole number and therefore unacceptable. Can someone please tell me what I am missing?

2. Dec 27, 2012

### Staff: Mentor

So if you got y as 1.5 then why don't you go back and double your other numbers to keep the balance and have integral values...

3. Dec 27, 2012

### v3ra

If I double those numbers, I will end up with the original a = 2, b = 4, x = 2, y = 3, z = 6 which would keep my formula unbalanced.

4. Dec 27, 2012

### Staff: Mentor

try picking a=30 ie 2*3*5 in other words w're adding in a new factor

Actually it seems a=2 would work too which means that the 12 should have worked fine.

Your mistake is in the 2a + 2b = 2z equation where it should have been a b instead of 2b

In the words of Shakespeare 2b or not 2b that is the question.

Last edited: Dec 27, 2012
5. Dec 28, 2012

### tahayassen

I'm curious: what course is this for?

6. Dec 28, 2012

### Staff: Mentor

Looks like HS Chemistry

7. Dec 28, 2012

### v3ra

I have 2a + b = 2z in that equation. 2a is referring to 2 hydrogen atoms in the first compound of the reaction which I have labeled 'a' and 'b' (literally '1b') is for the single hydrogen atom in the second compound of the reaction. 2z is of course the 2 hydrogen atoms in the last compound of the product.

8. Dec 28, 2012

### Staff: Mentor

Here's your work from the original post:

Code (Text):

(1)
2a + 2b = 2z
2(12) + 2(24) = 2z
24 + 48 = 2z
72 = 2z
z = 36

It should be 2a+b=2z right? and hence z=24

9. Dec 28, 2012

### v3ra

Ah, yes. This headache was the result of a typo..