Question

dm)
P. Calculate the mass of potassium
chlorate required to liberate 6.72 dm' of
oxygen at STP. Molar mass of KCIO, is
122.5 g mol-'.​

Answer 1

Answer:

24.5 grams of potassium chlorate

Explanation:

According to avagadro's law 1 mole of every substnce occupy 22.4L at STP and contain avagadro's number 6.023×10^23 of particles.

2KCLO3(s)------>2KCL(s)+3O2(g)

According to stachiometry:

2 moles of KCLO3 produce 3 moles of O2

Thus 3×22.4L=67.2L of O2 is produced from 2×122.5=245g of KCLO3

6.72L of O2 is produced from = FRAC24567.2×6.72=24.5g of KCLO3

This mass of potassium chlorate required to liberate 6.72dm^3 of oxygen at STP will be 24.5

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