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Chemistry

2KClO3 →MnO2\xrightarrow{\text{MnO}_2} 2KCl + 3O2

(i) Calculate the mass of KClO3 required to produce 6.72 lit of O2 at s.t.p.

[K = 39, Cl = 35.5, O = 16]

(ii) Calculate the number of moles of O2 in the above volume and also the number of molecules.

Stoichiometry

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Answer

2KClO3→MnO22KCl+3O22[39+35.52[393(22.4)+3(16)]+35.5]=67.2 lit.=245 g=149g\begin{matrix} 2\text{KClO}3 & \xrightarrow{MnO2} & 2\text{KCl} & + & 3\text{O}_2 \ 2[39 + 35.5 & & 2[39 & & 3(22.4) \ + 3(16)] & & + 35.5] & & = 67.2 \text{ lit.} \ = 245 \text{ g} & & = 149 \text{g} \ \end{matrix}

67.2 lit. of O2 is produced by 245 g of KClO3

∴ 6.72 lit of O2 will be obtained from 24567.2\dfrac{245}{67.2} x 6.72 = 24.5 g.

(ii) 22.4 lit = 1 mole
∴ 6.72 lit = 122.4\dfrac{1}{22.4} x 6.72 = 0.3 moles

1 mole = 6.023 x 1023 molecules
∴ 0.3 moles = 0.3 x 6.023 x 1023 molecules.

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