20.2 g of potassium nitrate is decomposed by heating according to the equation given below:
2KNO3 ⟶ 2KNO2 + O2
Calculate the following:
(a) Volume of oxygen gas obtained at STP.
(b) Mass of KNO2 formed.
[At. wt. of K = 39, N=14, O=16]
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$$\begin{matrix} 2\text{KNO}_3 & \longrightarrow & 2\text{KNO}_2 & + & \text{O}_2 \\ 2 \text{ moles} && 2 \text{ moles} && 1 \text{ mole} \\ 2[39 + 14 + 3(16)] && 2[39 + 14 + 2(16)] && 2(16) \\ = 202 \text{ g} && = 170 \text{ g} && = 32 \text{ g} \\ \end{matrix}$$
(a) From the equation, 202 g of KNO3 gives 22.4 L of O2 at STP.
∴ 20.2 g of KNO3 gives $\frac{22.4}{202}$ × 20.2 = 2.24 L of O2 at STP.
(b) From the equation, 202 g of KNO3 gives 170 g of KNO2.
∴ 20.2 g of KNO3 gives $\frac{170}{202}$ × 20.2 = 17 g of KNO2.
Hence, 2.24 L of oxygen is obtained at STP and 17 g of KNO2 is formed.