Question

What volume of oxygen will be produced by heating 245 kg of pure KClO3 at STP ? (Atomic weight of K = 39, Cl = 35.5 and O = 16)
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Answer 1

Answer:

67162.5‬ cubic decimetre (dm3) of O2

Explanation:

Step 1:

Write a balance Chemical equation for the following Reaction

KClO3 ---> KCl + 3/2O2

Step 2:

Calculating moles of KClO3:

Molar mass of KClO3 = 122.55 gm

Given Mass = 245 Kg

Moles of KClO3 = 245 / 122.55

Moles of KClO3 = 1.99 Kmol

Step 3:

According to balance Chemical Equation

1 Mole of KClO3 gives 3/2 moles of O2

1.99 KMole(or 1990 moles) gives "x" moles of O2

By balancing for x, we get

x = (3/2)*1990 moles of O2

x = 2985 moles of O2

Step 4:

Calculating Volume of O2 at STP

V = 2985*22.5 dm3

V= 67162.5‬ dm3