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Molecular oxygen is produced by the decomposition of KClO3. How many liters of O2 are produced by the complete decomposition of 100.0 g of KClO3 at STP?

User Mayowa
by
7.7k points

1 Answer

5 votes

Answer: 27.5 liters

Step-by-step explanation:

To calculate the moles, we use the equation:


\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

For
KClO_3

Given mass = 100.0 g

Molar mass of
KClO_3 = 122.5 g/mol

Putting values in above equation, we get:


\text{Moles of}KClO_3 =(100)/(122.5)=0.82moles


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

2 moles of
KClO_3 produces 3 moles of
O_2

0.82 moles of
KClO_3 produces =
(3)/(2)* 0.82=1.23 moles of
O_2

Volume of
O_2=moles* {\text {molar volume}}=1.23* 22.4=27.5L

Thus 27.5 liters of
O_2 are produced by the complete decomposition of 100.0 g of
KClO_3 at STP

User Adam Eisfeld
by
8.6k points
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