Question 1(2013)2KClO3 →MnO2\xrightarrow{ ext{MnO}_2}MnO2 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.

Question

Question 1(2013)2KClO3 →MnO2\xrightarrow{	ext{MnO}_2}MnO2​​ 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.

Accepted Answer

2KClO3→MnO22KCl+3O22[39+35.52[393(22.4)+3(16)]+35.5]=67.2 lit.=245 g=149g\begin{matrix} 2	ext{KClO}_3 & \xrightarrow{MnO_2} & 2	ext{KCl} & + & 3	ext{O}_2 \ 2[39 + 35.5 & & 2[39 & & 3(22.4) \ + 3(16)] & & + 35.5] & & = 67.2 	ext{ lit.} \ = 245 	ext{ g} & & = 149 	ext{g} \ \end{matrix}2KClO3​2[39+35.5+3(16)]=245 g​MnO2​​​2KCl2[39+35.5]=149g​+​3O2​3(22.4)=67.2 lit.​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}67.2245​ x 6.72 = 24.5 g.(ii) 22.4 lit = 1 mole∴ 6.72 lit = 122.4\dfrac{1}{22.4}22.41​ x 6.72 = 0.3 moles1 mole = 6.023 x 1023 molecules∴ 0.3 moles = 0.3 x 6.023 x 1023 molecules.

Stoichiometry — Percentage Composition — Empirical & Molecular Formula — Chemical Equations — Question 24

Question 1(2013)