8.2 grams of calcium nitrate is decomposed by heating according to the equation

2Ca (NO₃)₂ ⟶ 2CaO +4NO₂ + O₂

Calculate the following:

(a) Volume of nitrogen dioxide obtained at STP

(b) Mass of CaO formed

(Atomic weights : Ca- 40, N-14, O-16)

$\begin{matrix} \text{2Ca}(\text{NO}$3)2 & \longrightarrow & 2\text{CaO}& +&4\text{NO}2 & +& \text{O}2 \ 2[(40) + 2(14) +6(16)] & & 2(40 + 16) & & 4(14)+8(16) & & 1 \text{mole} \ = 328 \text{g} & & = 112\text{g} & & =184 & & =32 \text{g} \ \end{matrix}

(a) 328 g of calcium nitrate liberated 4 x 22.4 lit of NO₂

∴ 8.2 g of calcium nitrate liberated 4 x 22.4 x $\dfrac{8.2}{328}$ = 0.224 lit of NO₂

(b) 328 g of calcium nitrate produces 2 x 56 g of CaO

∴ 8.2 g of calcium nitrate will produce 2 x 56 x $\dfrac{8.2}{328}$ = 2.8 g of CaO.