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Chemistry

What volume of oxygen at s.t.p. will be obtained by the action of heat on 20 g KClO3 [K = 39, Cl = 35.5, O = 16]

Stoichiometry

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Answer

2KClO3 Δ 2KCl+3O22[39+35.53×22.4+48]=245=67.2\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 & & & & = 67.2 \ \end{matrix}

245 g of KClO3 gives 67.2 lit of O2

∴ 20 g of KClO3 will give 67.2245\dfrac{67.2}{245} x 20 = 5.486 lit of O2

Hence, 5.486 lit of oxygen will be obtained

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