1.56 g of sodium peroxide reacts with water according to the following equation:

2Na₂O₂ + 2H₂O ⟶ 4NaOH + O₂

Calculate:

(a) mass of sodium hydroxide formed,

(b) volume of oxygen liberated at S.T.P.

(c) mass of oxygen liberated.

$\begin{matrix} 2\text{Na}$2\text{O}2 & + \space 2\text{H}2\text{O} \longrightarrow & 4\text{NaOH}& & + & \text{O}2 \ 2[2(23) + 2(16)] & & 4(23 + 16 + 1) & & & 1 \text{mole} \ 156 \text{g} & & 160\text{g}& & &32 \text{g} \ \end{matrix}

(a) 156 g of sodium peroxide produces 160 g of sodium hydroxide

∴ 1.56 g of sodium peroxide will produce $\dfrac{160}{156}$ x 1.56

= 1.6 g of sodium hydroxide

(b) 156 g of sodium peroxide produces 22.4 L of oxygen

∴ 1.56 g of sodium peroxide will produce $\dfrac{22.4}{156}$ x 1.56

= 0.224 L

Converting L to cm³

As 1 L = 1000 cm³

So, 0.224 L = 224 cm³

(c) 156 g of sodium peroxide produces 32 g of oxygen

∴ 1.56 g of sodium peroxide will produce $\dfrac{32}{156}$ x 1.56 = 0.32 g