Answer :
I'll give you the general method based on the example you provided :
n Fe +m H2S04 ---> p Fe2(SO4)3 +q H2
We want to find n,m,p and q.
The idea it to make sure the conservation of the elements happens. If you have 3 Oxygen on the left side, you need to have 3 Oxygen on the right side.
Here we have :
Elements | Left side (reagents) : | Right side (products) :
-----------------------------------------------------------------------------------
Fe n 2p
H 2m 2q
S m 3p
O 4m 4*3p
We want m,n,p,q to be as small as possible so let's start by trying m=1:
n Fe +H2S04 ---> p Fe2(SO4)3 +q H2
From that we can see that:
* Conservation of Oxygen, Sulfur : p=3
That gives us n Fe +3H2S04 ---> Fe2(SO4)3 +H2
The conservation of Hydrogen thus gives :
n Fe +3H2S04 ---> Fe2(SO4)3 +3H2
The conservation of Fe then yields n=2
[tex]2Fe +3H_2S0_4 \rightarrow Fe_2(SO_4)_3 +3H_2[/tex]
n Fe +m H2S04 ---> p Fe2(SO4)3 +q H2
We want to find n,m,p and q.
The idea it to make sure the conservation of the elements happens. If you have 3 Oxygen on the left side, you need to have 3 Oxygen on the right side.
Here we have :
Elements | Left side (reagents) : | Right side (products) :
-----------------------------------------------------------------------------------
Fe n 2p
H 2m 2q
S m 3p
O 4m 4*3p
We want m,n,p,q to be as small as possible so let's start by trying m=1:
n Fe +H2S04 ---> p Fe2(SO4)3 +q H2
From that we can see that:
* Conservation of Oxygen, Sulfur : p=3
That gives us n Fe +3H2S04 ---> Fe2(SO4)3 +H2
The conservation of Hydrogen thus gives :
n Fe +3H2S04 ---> Fe2(SO4)3 +3H2
The conservation of Fe then yields n=2
[tex]2Fe +3H_2S0_4 \rightarrow Fe_2(SO_4)_3 +3H_2[/tex]