Number 1: answer is Zn + 2HCl ----> ZnCl2 + H2
This one was a bit simple
we have:
zn- 1 zn-1
H-1 H-2
Cl-1 Cl-2
After this you can see that both H and Cl need to be multiplied by 2 on the reactants side. By putting 2HCl, you have accomplished this goal
Number 2: 4Fe + 3O2 ----> 2Fe2O3
We have:
Fe- 1 Fe-2
O-2 O-3
So unlucky for us, none of these were already balanced. I started first with Fe and started plugging in numbers. So if i were to put a 2 in front of Fe, it would cause 2-Fe and then 2-Fe on the other side, making it balanced.
The problem is that the O values still need to be balanced. If you notice, both Os have a 3 or a 2. 3 x 2= 6. SO you would need to add the opposite number on each. (3O2).
Number 3: 2H2O2 -----> 2H2O + O2
The answer is balanced as it has
H- 4 H-4
O-4 O-4
To figure this out, you would have to first look at what you need
The H values already look balanced, so we will look at O
There are 2 on the reactant side and 3 on the product
Start by adding a 2 onto H2O2 (2H2O2)
We now have 4 H and 4 O values on that side, but we still need to change the products side
By adding a 2 to H2O, You would then have the number of required values.
