The decomposition of potassium chlorate yields oxygen gas. If the yield is 95%,

Question

The decomposition of potassium chlorate yields oxygen gas. If the yield is 95%, how many grams of KClO are needed to produce 10.0 L of O?
2KClO(s) 2KCl(s) + 3O(g)
The decomposition of potassium chlorate yields oxygen gas. If the yield is 95%, how many grams of KClO are needed to produce 10.0 L of O?
2KClO(s) 2KCl(s) + 3O(g)
The decomposition of potassium chlorate yields oxygen gas. If the yield is 95%, how many grams of KClO are needed to produce 10.0 L of O?
2KClO(s) 2KCl(s) + 3O(g) The decomposition of potassium chlorate yields oxygen gas. If the yield is 95%, how many grams of KClO are needed to produce 10.0 L of O?
2KClO(s) 2KCl(s) + 3O(g)

Explanation / Answer

The reaction taking place is:

2KClO3 ---> 2KCl + 3O2

1 mole of every gas as STP occupies 22.4 L of volume.

Assuming STP conditions, 10 L of volume is occupied by: (1/22.4)*10 = 0.446 moles of gas

As per the reaction stoichiometry,

1 mole of KClO3 produces 1.5 moles of O2

Since the yield is 95%, so

1 mole of KClO3 produces : 1.5*0.95 = 1.425 moles of O2

So,

0.446 moles of O2 will be produced by: (1/1.425)*0.446 = 0.313 moles KClO3

So,

Mass of KClO3 required = Moles*MW = 0.313*122.55 = 38.36 g

Hope this helps !

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