Question

Calculate the mass of potassium chlorate required to liberated 6.72dm^3 of oxygen at STP molar mass of potassium chlorate is 112.5g/mol

Answer 1

Given Data :-

Volume of oxygen = 6.72 dm^3

Molar mass of potassium chlorate = 112.5 g/mol

Moles of oxygen = Volume of oxygen / volume at STP

Moles of oxygen = 6.72 /22.4

Moles of oxygen = 0.3 mol

BALANCED CHEMICAL EQUATION :-

2 KClO3 <-----> 2KCl + 3O2

3 moles of O2 produces 2 moles of KCl

Then

0.3 moles of O2 = 2/3 × 0.2 moles of KClO3

0.3 moles of O2 = 0.2 moles of KClO3

Now

Moles of KClO3 = mass/ molar mass

Mass of KClO3 = moles × molar mass

Molar mass of KClO3 = 39 + 35.5 +16×3= 122.5g/mol

Mass of KClO3 = 0.2 × 122

Mass of KClO3 = 24.5g

Hence mass of potassium chlorate is 24.5g

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