Question

Balance the chemical equation: $\mathrm{Al}\left(\mathrm{s}\right)+{\mathrm{O}}_2\left(\mathrm{g}\right)\to {\mathrm{Al}}_2{\mathrm{O}}_3\left(\mathrm{s}\right)$

Answer 1

The skeletal chemical equation: $\mathrm{Al}\left(\mathrm{s}\right)+{\mathrm{O}}_2\left(\mathrm{g}\right)\to {\mathrm{Al}}_2{\mathrm{O}}_3\left(\mathrm{s}\right)$

Step 1: As in the balanced chemical equation, the number of elements in a compound each on both the sides should have to equal. Basically in a balanced chemical equation number of elements on ($\mathrm{LHS}$) should be equal to that of the elements on ($\mathrm{RHS}$).

Element	Reactant side (LHS)	Product side (RHS)
Aluminium	1	2
Oxygen	2	3

Step 2: So for balancing all the atoms on both the sides, firstly balance $\mathrm{Al}$ on the $\mathrm{LHS}$ by multiplying it by 2 then for balancing the $\mathrm{O}$ atoms multiply $3\mathrm{to}{\mathrm{O}}_2$ on $\mathrm{LHS}$

Then further at last, just multiply the whole equation with 2.

Element	Reactant side (LHS)	Product side (RHS)
Aluminium	4	4
Oxygen	6	6

Therefore, the reaction is : $4\mathrm{Al}\left(\mathrm{s}\right)+3{\mathrm{O}}_2\left(\mathrm{g}\right)\to 2{\mathrm{Al}}_2{\mathrm{O}}_3\left(\mathrm{s}\right)$