Are you familiar with matrices or simultaneous equations? If so, this would make it much easier for you. If not, ill go through it in a different way

 

I at least know simultaneous equations, I can't remember if I know matrices or not.

 

Here you are, Lycel
Allright, theres 2 ways to solve this. Theres a "traditional" way and an algebraic way. While the traditional way is easier for straightforward equations, the algebraic way will be a hell of a lot better for bigger equations. The Algebraic way works for any sized equation and is not a try/test way.
Traditional way:
H2SO4 + Pb(OH)4 --> Pb(SO4)2 + H2O
We start with the Sulfate ions because we want to leave Hydrogen and Oxygen to the end.

So we count the number of Sulfates on the left side and the number of Sulfates on the right hand side of the equation.
Left: 1x (SO4) Right: 2x(SO4)
So we immediately know that the sulfuric acid will have a coefficient of 2 or more. Lets chuck the 2 in there.
2H2SO4 + Pb(OH)4 --> Pb(SO4)2 + H2O

Now lets count the Lead atoms.
Left: 1 Lead Right: 1 Lead
These are balanced so lets not mess with them for now.

Now lets balance the Hydrogens.
Left: 8 Hydrogens Right: 2 Hydrogens
So we see that the hydrogens are unbalanced. The left side has 4 times more hydrogen atoms. So we times the H2O by 4 to get a total of 8 Hydrogen atoms on both sides of the equation.
2H2SO4 + Pb(OH)4 --> Pb(SO4)2 + 4H2O

Now lets check the last atoms (Oxygen).
Left: 12 Oxygens Right: 12 Oxygens
Oh Look! They're all balanced
So you end up with the final equation of:

2H2SO4 + Pb(OH)4 --> Pb(SO4)2 + 4H2O

Algebraic Way:
H2SO4 + Pb(OH)4 --> Pb(SO4)2 + H2O
The way's method is to form a simultaneous equation for each type of atom.
Step 1: Bring everything from the equation over to one side (Left)
H2SO4 + Pb(OH)4 --> Pb(SO4)2 + H2O
turns to:
H2SO4 + Pb(OH)4 - Pb(SO4)2 -H2O = 0 (Dont overlook the minus signs, they're important!)

Step 2: Write every type of atom vertically below each other.
Pb:
S:
O:
H:
Each of these atoms will form one equation (Hence the : )

Now, give every compound involved a name. Its best to call them x1, x2, x3, x4... you can give them other names aswell like x, y, z, a...
I'll use x, y, z, a to make it less confusing further on. These you write horizontally:




Now comes the tricky part. You have to see how many of each atom in each compound.
Eg. Pb(OH)4
There is 1 Pb atom, 4 Oxygens and 4 Hydrogen atoms. And as Pb(OH)4 is = x2 you write the number of atoms in the x2 column as I will do now.



Now we do the same things for the other compounds, however, the values of - Pb(SO4)2, -H2O will be Negative.
H2SO4 + Pb(OH)4 - Pb(SO4)2 -H2O = 0

So you should end up with this:


So from this we can form 4 simultaneous equations
You go across, and multiply the numbers by the x variable above them. Eg. The first equation will be:

Pb: 0x + 1y - 1z +0a = 0 Which is Pb: 1y - 1z = 0
S: 1x + 0y - 2z + 0a = 0 S: 1x - 2z = 0
O: 4x + 4y -8z - 1a = 0 O: 4x + 4y - 8z - 1a = 0
H: 2x + 4y - 0z - 2a = 0 H: 2z + 4y - 2a = 0

So if we solve these 4 equations we get the values of x, y, z, a.
x = the coefficient for the H2SO4
y = the coefficient for Pb(OH)4
z = the coefficient for Pb(SO4)2
a = the coefficient for H2O

These would be really simple to solve with matrices, but I'm really bad at drawing these on here so I'll use an online solver.
Gyazo - 0ae53d9d0dd1fd4d6570f75f5da914d5.png
Gyazo - 4c4d5b5b121c18f1ae5d3f138b16c35b.png
Gyazo - d341d7daa97d39adbba5df5b95c823a0.png
Gyazo - 3ffd4508f1951563d0d0fcda269252ba.png
Gyazo - 9cf624ae4fc2e4662027383807104f25.png
Gyazo - 8b044aae17413f29576da53c96fc5c1c.png

The last page shows us that
x1 = 1/2 = x
x2 = 1/4 = y
x3 = 1/4 = z
x4 = Free = a This means that x4 or a is determined by the other 3 variables.
So now lets put these coefficients into our equation:
(1/2)H2SO4 + (1/4)Pb(OH)4 --> (1/4)Pb(SO4)2 + H2O
So we times the whole equation by the biggest denominator to get whole numbers which is 4.
4 * (1/2) * H2SO4 + 4 * (1/4) * Pb(OH)4 --> 4 * (1/4) * Pb(SO4)2 + 4 * H2O

So our equation will end up as:
2H2SO4 + Pb(OH)4 --> Pb(SO4)2 + 4H2O



So that's it. Algebraic way is much more useful in higher level chemistry, but if you know that way, you'll be your classmates way ahead
I hope this helped. If its still unclear, tell me and I'll try to simplify it even more
Phew, this took me a whole hour to write xD

 

Oh my gosh you have no idea how much this has helped 3
I'm clueless in Chemistry at the moment because they failed to teach this stuff to us in general science from year 8-10... but now I understand at least this! ^-^
I didn't know you were such a nerd Phil :3 hehe.
Thank you sososososooo much~

 

Hahahaha no worries, it would be a shame for me to learn all that stuff but to never pass it on

 

Lycel look for any giveaways when using the traditional way, eg Polyatomic ions (SO4, NO3)

 

If anyone has more maths problems, feel free to hit me up ^^

 

GTechNerd do you know the algebraic way to balance equations?

 

Nah. I just do them in my head.

 

Mooooore Maaaaaths!!!!!! FEEED ME MATHS!

 

Fine, here's one:

Find the exact solution of the equation 9^2x = 27^(1-x). Have fun.

 

Thanks
9^(2x) = 27^(1-x)
=> 3^(2(2x)) = 3^(3(1-x))
=> 4x = 3-3x
=> 7x = 3
=> x = 3/7

Lol I resorted to logarithms first but then I realised the common factors ^^

 

Im in eighth grade and my science teacher is making me do high school physics if i have any questions ill come to you

 

Sweet as