Balance the following chemical equation . (a). ZnS+O2-ZnO+SO2

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Solution

Left side, we have 1 $O_{2}$ , ie 2 O. Right, we need 3 O.

This is an odd-even condition, ie odd number on one side and even on the other. So we convert both to odd, by putting 2ZnO on right.
So we get
$Z n S + O_{2} ⟶ 2 Z n O + S O_{2}$
Now, balance Zn on left by putting 2Zn
$2 Z n S + O_{2} ⟶ 2 Z n O + S O_{2}$
Now , balance S on right by putting 2 $S O_{2}$ .
$2 Z n S + O_{2} ⟶ 2 Z n O + 2 S O_{2}$
Now, only O is unbalanced.
Right we have 6 O, left ee to have only 2. So multiply O2 by 3, to get 6 on both sides
$2 Z n S + 3 O_{2} ⟶ 2 Z n O + 2 S O_{2}$

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