Calculate the volume of oxygen at S.T.P. obtained by heating 35gm kclo3 [at wt k=39,cl=35.5,o=16]
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Considering the equation,
2KClO3 -->KCl + 3O2
Molar mass of KClO3 = 122.5g
At STP,
1 mole of KClO3 occupies 22.4 litres
that is 122.5g of KClO3 occupies 22.4 liters
12.25 g of KClO3 occupies 2.24 litres
as according to avogadro's law, when gases combine, they do so in volumes bear a constant ratio to each other, so
2 volume KClO3 produces 3 Volume of O2
2.24 liters of KClO3 will produce [tex]2.24*\frac{3}{2}[\tex] liter of O2 that equals 3.36 liters
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