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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$$\begin{matrix} 2\text{Ca}(\text{NO}_3)_2 & \longrightarrow & 2\text{CaO} & + & 4\text{NO}_2 & + & \text{O}_2 \\ 2[(40) + & & 2(40 & & 4[22.4] \\ 2(14 +3(16))] & & + 16) & & \text{ lit.} \\ = 328 \text{ g} & & = 112 \text{g} \\ \end{matrix}$$
(i) 328 g of Ca(NO3)2 produces 4(22.4) lit of nitrogen dioxide.
∴ 8.2 g of Ca(NO3)2 will produce $\frac{4 \times 22.4}{328}$ 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 $\frac{112}{328}$ x 8.2 = 2.8 g of calcium oxide.
Hence, 2.8 g of calcium oxide is produced.