$(i)$ What mass of hydrogen peroxide will be present in $2 \, L$ of a $5 \, M$ solution?
$(ii)$ Calculate the mass of oxygen which will be liberated by the decomposition of $200 \, mL$ of this solution.

(N/A) $(i)$ Molar mass of $H_2O_2 = 34 \, g \, mol^{-1}$.
$1 \, L$ of $5 \, M$ solution of $H_2O_2$ contains $34 \times 5 = 170 \, g$ of $H_2O_2$.
Therefore,$2 \, L$ of $5 \, M$ solution of $H_2O_2$ contains $170 \times 2 = 340 \, g$ of $H_2O_2$.
$(ii)$ $200 \, mL$ $(0.2 \, L)$ of $5 \, M$ solution contains $H_2O_2 = 5 \, mol \, L^{-1} \times 0.2 \, L = 1 \, mol$.
Mass of $1 \, mol$ of $H_2O_2 = 34 \, g$.
The decomposition reaction is: $2H_2O_2 \rightarrow 2H_2O + O_2$.
From the stoichiometry,$2 \, mol$ of $H_2O_2$ $(68 \, g)$ produces $1 \, mol$ of $O_2$ $(32 \, g)$.
Thus,$1 \, mol$ of $H_2O_2$ $(34 \, g)$ will produce $\frac{32}{2} = 16 \, g$ of $O_2$.