0.1 mole of potassium permanganate was heated at $300^{\circ} \mathrm{C}$. What is the weight (ing) of the…

0.1 mole of potassium permanganate was heated at $300^{\circ} \mathrm{C}$. What is the weight (ing) of the residue? $(\mathrm{Mn}=55 \mathrm{u} ; \mathrm{K}=39 \mathrm{u} ; \mathrm{O}=16 \mathrm{u})$
  1. 14.2
  2. 1.6
  3. 15.8
  4. 7.1

Solution

$2 \mathrm{KMnO}_4 \xrightarrow{300^{\circ} \mathrm{C}} \underbrace{\mathrm{K}_2 \mathrm{MnO}_4+\mathrm{MnO}_2}_{\text {Solid residue }}+\mathrm{O}_2$ Molecular weight of $\mathrm{KMnO}_4=158 \mathrm{u}$ Mole $=0.1$ $\begin{aligned} & \text { Weight }=\text { Molecular weight } \times \text { mole } \\ & =0.1 \times 158 \\ & \text { Weight of Total } \mathrm{KMnO}_4=15.8 \mathrm{~g} \\ & \begin{array}{l}\text { After heat } \mathrm{O}_2 \text { will evolved }\end{array} \\ & \begin{aligned} \therefore \quad \text { Weight of } \mathrm{O}_2=\text { mole } \times \text { molecular weight } \\ =0.1 \times 16\end{aligned} \\ & \begin{aligned} \therefore \quad & 1.6 \mathrm{~g}\end{aligned} \\ & \begin{aligned} \text { Weight of Residue } & =15.8-1.6 \\ & =14.2 \mathrm{~g}\end{aligned}\end{aligned}$

Asked in: AP EAMCET 2024 (21 May Shift 1)

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