Question 51(b)O2 is evolved by heating KClO3 using MnO2 as a catalyst.2KClO3 →MnO2\xrightarrow{ ext{MnO}_2}MnO2 2KCl + 3O2(i) Calculate the mass of KClO3 required to produce 6.72 litre of O2 at STP.(atomic masses of K = 39, Cl = 35.5, 0 = 16).(ii) Calculate the number of moles of oxygen present in the above volume and also the number of molecules.(iii) Calculate the volume occupied by 0.01 mole of CO2 at STP.

Question

Question 51(b)O2 is evolved by heating KClO3 using MnO2 as a catalyst.2KClO3 →MnO2\xrightarrow{	ext{MnO}_2}MnO2​​ 2KCl + 3O2(i) Calculate the mass of KClO3 required to produce 6.72 litre of O2 at STP.(atomic masses of K = 39, Cl = 35.5, 0 = 16).(ii) Calculate the number of moles of oxygen present in the above volume and also the number of molecules.(iii) Calculate the volume occupied by 0.01 mole of CO2 at STP.

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.​(i)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 = 1.806 x 1023 molecules.(iii)Volume occupied by 1 mole of CO2 at STP = 22.4 litresSo, volume occupied by 0.01 mole of CO2 at STP= 22.4 × 0.01= 0.224 litres

Mole Concept & Stoichiometry Miscellaneous Exercises — Question 66

Question 51(b)