If 36 gm of $\mathit{Mg}$ reacts with $O_2$ to give $2\mathit{MgO}$ in the ratio of 3:2 then what is the mass of $O_2$ in grams that is required for the reaction?

If 36 gm of $\mathit{Mg}$ reacts with $O_2$ to give $2\mathit{MgO}$ in the ratio 3:2 then the mass of $O_2$ required is 24 gm.
The mass of the Mg is 36 gm and let the mass of $O_2$ be “$2X$”.
Therefore, the reaction takes place as follows:
$\mathit{Mg }+ O_2 \to 2\mathit{MgO}$
So, according to the law of constant proportions, 3 units of $\mathit{Mg}$ will react with 2 units of $O_2$ to give $2\mathit{MgO}$.
Therefore, 3 units of $\mathit{Mg}$ is 36gm so 2 units of $O_2$ will be   

$\Rightarrow 36\mathit{gm}= 3X$
$\Rightarrow \mathit{ X }= \frac{36}{3}\mathit{gm}$

$\Rightarrow \mathit{ X }= 12 \mathit{gm}$

Therefore, the amount of required oxygen $=$ $2X$
$\Rightarrow 2\times 12$
$\Rightarrow 24 \mathit{gm}$