Question

Calculate the mass of potassium chlorate required to liberate 6.72 dm cube of oxygen at stp

Answer 1

Molar mass of KClO3 is roughly 122.5 g/mol.

KClO3===∆===> KCl + 3/2 O2.

So 122.6 g of KClO3 will liberate 1.5 mole of oxygen or 22.4 x 1.5= 33.6 Litres of oxygen at STP.

Therefore for the liberation of 6.72 g of oxygen at STP required amount of KClO3 =(6.72 x 122.6)/33.6.

= 24.52 g.

Answer 2

Answer:

It is the milae mass of potassium chlorate required to liberate 6.72 dm cube of oxygen at stp......