O₂ is evolved by heating KClO₃ using MnO₂ as a catalyst

$2 KlO subscript 3 space rightwards arrow with MnO subscript 2 space space space space on top space 2 KCl space plus space 3 straight O subscript 2$

i) Calculate the mass of KClO₃ required to produce 6.72 litres of O₂ at STP.

[atomic masses of K =39, Cl=35.5, O=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

2 KClO subscript 3 space space rightwards arrow with space space space space MnO subscript 2 space space space space space space on top space 2 KCl space space space space plus 3 straight O subscript 2 Molecular space mass space of space space 2 KClO subscript 3 space equals 2 space straight x space left parenthesis 39 plus 35.5 plus 3 straight x 16 right parenthesis space equals space 2 space straight x space 122.5 space equals 245 space straight g 2 KCl space equals 2 space left parenthesis 39 space plus 35.5 right parenthesis space equals 149 space straight g 3 straight O subscript 2 space equals space 3 space straight x space 22.4 space equals 67.2 space straight l 67.2 space litre space of space straight O subscript 2 space produced space by space KClO subscript 3 space equals 245 space straight g space 1 space litre space of space straight O subscript 2 space produced space by space KClO subscript 3 space equals fraction numerator 245 over denominator 67.2 end fraction 6.72 space of space straight O subscript 2 space produced space by space KClO subscript 3 space equals fraction numerator 245 space straight x space 6.72 over denominator 67.2 end fraction space space space space space space space space space space space space space space space space space space space space space space space equals space 24.5 space straight g space ii right parenthesis space space 245 space straight g space of space KClO subscript 3 space contains space equals space 3 space moles space of space oxygen space space space space 24.5 space straight g space of space KClO subscript 3 space contains space equals fraction numerator 3 space straight x 24.5 over denominator 245 end fraction space space space space equals 0.3 space mole since space comma space 1 space mole space of space oxygen space contain space equals 6.022 space straight x 10 to the power of 23 space molecules therefore comma 0.3 space mole space of space oxygen space contain space equals 6.022 space straight x 10 to the power of 23 space straight x 0.3 space space space space space space equals space 1.8066 space straight x space 10 to the power of 23 space molecules

iii right parenthesis space 1 space mole space of space CO subscript 2 space occupied space equals space 22.4 space straight l 0.01 space mole space of space CO subscript 2 space occupied space equals space 22.4 space straight x space 0.01 space space space space space space space space space space space space space space space space equals space 0.224 space straight l

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