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} 2\text{Ca}(\text{NO}$3)2 & \longrightarrow & 2\text{CaO} & + & 4\text{NO}2 & + & \text{O}2 \ 2[(40) + & & 2(40 & & 4[22.4] \ 2(14 +3(16))] & & + 16) & & \text{ lit.} \ = 328 \text{ g} & & = 112 \text{g} \ \end{matrix}

(i) 328 g of Ca(NO₃)₂ produces 4(22.4) lit of nitrogen dioxide.

∴ 8.2 g of Ca(NO₃)₂ will produce $\dfrac{4 \times 22.4}{328}$ x 8.2 = 2.24 lit of nitrogen dioxide.

Hence, vol of nitrogen dioxide evolved = 2.24 lit.

(ii) 328 g of Ca(NO₃)₂ produces 112 g of calcium oxide.

∴ 8.2 g of Ca(NO₃)₂ will produce $\dfrac{112}{328}$ x 8.2 = 2.8 g of calcium oxide.

Hence, 2.8 g of calcium oxide is produced.