Question

calculate the volume of oxygen at STP liberated by heating 12.25 gram of kClO3?

Answer 1

The chemical equation for decomposition of KClO3

can be written as:

2KClO3→2KCl+3O2

245 g of KClO3

gives 67.2

L of O2

12.25

g of KClO3

gives 67.2245×12.25

=3.36 L

Answer 2

336 L of oxygen at STP are liberated by heating 12.25 gram of $KClO_3$

Explanation:

According to avogadro's law, 1 mole of every substance occupies 22.4 Liters at STP and contains avogadro's number $(6.023\times 10^{23})$  of particles.

$2KClO_3(s)\rightarrow 2KCl(s)+3O_2(g)$

moles of $KClO_3=\frac{\text {given mass}}{\text {Molar mass}}=\frac{12.25g}{122.5g/mol}=10moles$

According to stoichiometry:

2 moles of $KClO_3$ produces = 3 moles of $O_2$

Thus 10 moles of $KClO_3$ produces = $\frac{3}{2}\times 10=15$ moles of $O_2$

1 mole of $O_2$ occupies = 22.4 L at STP

Thus 15 mole of $O_2$ occupies = $\frac{22.4}{1}\times 15=336 L$ at STP

Thus 336 L of oxygen at STP are liberated by heating 12.25 gram of $KClO_3$

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