How much potassium chlorate should be heated to produce 2024 L of oxygen at STP?
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When potassium chlorate is heated, then following reaction takes place -
2KClO3 −→heat
2KCl + 3O2
As we know, 1 mole of oxygen gas occupies 22.4 L of volume
Let x moles of oxygen gas will occupy 2.24 L
then x = 2.2422.4 = 0.1 moles2.24 ×122.4 = 0.1 moles
Thus, 2.24 L of volume has 0.1 moles of oxygen gas.
Now, look at the reaction,
2 moles of KClO3 is giving 3 moles of oxygen
Let y of KClO3 is giving 0.1 moles of oxygen
Then y = 0.1×230.1×23 = 0.0667 moles
Number of moles = mass of KClO3/ Molar mass
Molar mass of KClO 3 = [ 39 + 35.5 + 3×16] =122.5 gmol-1
0.0667 = mass of KClO3 / 122.5
mass of KClO3 = 8.171 g
2KClO3 −→heat
2KCl + 3O2
As we know, 1 mole of oxygen gas occupies 22.4 L of volume
Let x moles of oxygen gas will occupy 2.24 L
then x = 2.2422.4 = 0.1 moles2.24 ×122.4 = 0.1 moles
Thus, 2.24 L of volume has 0.1 moles of oxygen gas.
Now, look at the reaction,
2 moles of KClO3 is giving 3 moles of oxygen
Let y of KClO3 is giving 0.1 moles of oxygen
Then y = 0.1×230.1×23 = 0.0667 moles
Number of moles = mass of KClO3/ Molar mass
Molar mass of KClO 3 = [ 39 + 35.5 + 3×16] =122.5 gmol-1
0.0667 = mass of KClO3 / 122.5
mass of KClO3 = 8.171 g
Nazeefa0007:
Hello and thank u for the answer
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