Question

What minimum moles of KClO3 must be decomposed , in order to produce at least 100 grams of KCl and at least 100 grams of grams of O2.
(Reaction involved:- 2KClO3→2KCl+3O2)

Answer 1

Explanation:

In order to be able to calculate the volume of oxygen gas produced by this reaction, you need to know the conditions for pressure and temperature.

Since no mention of those conditions was made, I'll assume that the reaction takes place at STP, Standard Temperature and Pressure.

STP conditions are defined as a pressure of

100 kPa

and a temperature of

0

∘

C

. Under these conditions for pressure and temperature, one mole of any ideal gas occupies

22.7 L

- this is known as the molar volume of a gas at STP.

So, in order to find the volume of oxygen gas at STP, you need to know how many moles of oxygen are produced by this reaction.

The balanced chemical equation for this decomposition reaction looks like this

2

KClO

3(s]

heat

×

−−−→

2

KCl

(s]

+

3

O

2(g]

↑

⏐

⏐

Notice that you have a

2

:

3

mole ratio between potassium chlorate and oxygen gas.

This tells you that the reaction will always produce

3

2

times more moles of oxygen gas than the number of moles of potassium chlorate that underwent decomposition.

Use potassium chlorate's molar mass to determine how many moles you have in that

231-g

sample

231

g

⋅

1 mole KClO

3

122.55

g

=

1.885 moles KClO

3

Use the aforementioned mole ratio to determine how many moles of oxygen would be produced from this many moles of potassium chlorate

1.885

moles KClO

3

⋅

3

moles O

2

2

moles KClO

3

=

2.8275 moles O

2

So, what volume would this many moles occupy at STP?

2.8275

moles

⋅

22.7 L

1

mol

=

64.2 L

The answer is rounded to three sig figs

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