Question

Oxygen the mass of potassium Chlorate required to produce 26.88 litres of oxygen as catalyst
2KCIO -----> 2KCl+3O2
(1) Calculate the mass of potassium Chlorate required to produce 26.88 litres of oxygen at STP. (K=39,Cl=35.5,O=16)
(2)Calculate the number of moles of oxygen present in the above volumes.​

Answer 1

we have to find

1. the mass of potassium chlorate required to produce 26.88 litres of oxygen at STP.
2. the number of moles of oxygen present in the above volumes.

solution : dissociation reaction of potassium chlorate is..

2KClO3 => 2KCl + 3O2

here you see, 2 moles of KClO3 produce 3 moles of oxygen.

molar mass of KClO3 = 122.5 g/mol

volume of 1 mole of gas at STP = 22.4 litres

so, 2 × 122.5 g of KCl produce 3 × 22.4 litres of oxygen gas at STP

so, 26.88 litres will be produced by 2 × 122.5/(3 × 22.4) × 26.88 g of KClO3

= 2 × 122.5/(3 × 22.4) × 26.88

= 98g

Therefore 98g of KClO3 required to produce 26.88 litres of oxygen gas at STP.

no of moles of oxygen gas = volume of oxygen gas/22.4

= 26.88/22.4

= 1.2

Therefore the no of moles of oxygen gas is 1.2 mol

Answer 2

Answer:

Hope it help you

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