Question

Find Oxidation number of S:
KAl(SO4)2.12H2O​

Answer 1

Answer: S=6

Explanation:

Constants K=1,Al=3,O=-2

Because there is no charge above the compound So the compound is Natural so it equals 0

KAL(SO4)2=0. Now we replace each known elements with his number

1+3+(S× -2×4)=0. We multiply all the elements inside the parenthesis by 2

1+3+(2S+8-×2)=0

1+3+2S+16- =0

We sum allthe numbers

4 + 16 - 2S =0

12-2S=0. move the variations to the right side

2S=12 by ÷2

S=6