(i) What is limiting reactant?
(ii) Oxygen is prepared by catalytic decomposition of potassium chlorate (KClO3
).
Decomposition of potassium chlorate gives potassium chloride (KCl) and oxygen (O2
).
If 2.4 mol of oxygen is needed for an experiment, how many grams of potassium chlorate must be
decomposed?
(At. mass of K = 39, Cl=35.5, O = 16)
Answers
Answered by
0
Answer:
Decomposition of KClO
3
takes place as:
2KClO
3
(s)→2KCl(s)+3O
2
(g)
2 mole of KClO
3
=3 mole of O
2
∵ 3 mole O
2
formed by 2 mole KClO
3
∴2.4 mole of O
2
will be formed by:
(
3
2
×2.4) mol KClO
3
= 1.6 mole of KClO
3
.
Mass of KClO
3
= Number of moles × Molar mass
=1.6×122.5=196g
hope it helps ☺️
Answered by
3
Answer:
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