how many moles of pottasium chlorate should decompose to give 67.2 litres of oxygen
Answers
Answer:
The reaction is 2KClO3→2KCl+3O2.
The decomposition of two moles of potassium chlorate gives three moles of oxygen at STP.
At STP, one mole of oxygen occupies 22.4 L.
Hence, 3 moles of oxygen will occupy 67.2 L.
Thus, to obtain 67.2 L of oxygen at STP, 2 moles of potassium chlorate should be decomposed.
Answer:
All you need in order to answer this question is the balanced chemical equation for the decomposition reaction of potassium chlorate
2
K
C
l
O
3
(
s
)
→
2
K
C
l
(
s
)
+
3
O
2
(
g
)
Notice that you have a
2
:
3
mole ratio between potassium chlorate and oxygen gas, which means that, regardless of how many moles of the former react, you'll always produce 3/2 times more moles of the latter.
Since you know how many moles of oxygen you need the reaction to produce, you can work backwards and determine how many moles of potassium chlorate you need by using the same mole ra
Explanation: