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Chemistry

2KClO3 ⟶ 2KCl + 3O2 ; C + O2 ⟶ CO2. Calculate the amount of KClO3 which on thermal decomposition gives 'X' vol. of O2, which is the volume required for combustion of 24 g. of carbon. [K = 39, Cl = 35.5, O = 16, C = 12].

Stoichiometry

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Answer

C+O2ΔCO212 g22.4 lit\begin{matrix} \text{C} & + & \text {O}2 & \xrightarrow{\Delta} & \text{CO}2 \ 12 \text{ g} & & 22.4 \text{ lit} & & \end{matrix}

12 g of C requires 22.4 lit of O2

∴ 24 g of C will require 22.412\dfrac{22.4}{12} x 24 = 44.8 lit. of O2

Hence, 24 g of C will require 44.8 lit of O2

2KClO3 Δ 2KCl+3O22[39+35.5(3×22.4)+48]=245 g=67.2 lit\begin{matrix} 2\text{KClO}3 & \xrightarrow{\space \Delta \space} & & 2\text{KCl} & + & 3\text{O}2 \ 2[39 + 35.5 & & & & & (3 \times 22.4) \ + 48] = 245 \text{ g} & & & & & = 67.2 \text{ lit} \ \end{matrix}

67.2 lit of O2 is produced by 245 g of KClO3

∴ 44.8 lit of O2 is produced by 24567.2\dfrac{245}{67.2} x 44.8 = 163.33 g of KClO3

Hence, amount of KClO3 is 163.33 g.

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