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)
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$$\begin{matrix} \text{2Ca}(\text{NO}_3)_2 & \longrightarrow & 2\text{CaO}& +&4\text{NO}_2 & +& \text{O}_2 \\ 2[(40) + 2(14) +6(16)] & & 2(40 + 16) & & 4(14)+8(16) & & 1 \text{mole} \\ = 328 \text{g} & & = 112\text{g} & & =184 & & =32 \text{g} \\ \end{matrix}$$
(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 $\frac{8.2}{328}$ = 2.24 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 $\frac{8.2}{328}$ = 2.8 g of CaO.