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| Post
Answer |
One method for preparing oxygen is to heat pottassium chlorate in the prescence of a catalyst. Find the STP volume of oxygen that can be produced by 5.64g KClO3.
The answer is supposed to be 1.59 L O2 but for some reason I keep getting 1.55
Help? |
| Answer |
I am sure the answer is 1.55L, not 1.59 L
The equation for this is
2KClO3 „³ 2 KCl + 3O2
first find the number of moles of KClO3
gm molar mass of KClO3 = 122 gm /mole
so 5.64 gm will contain 5.64/122.6 = 0.0462 moles
according to the equation every 2 moles of KClO3 will release 3 moles of O2
So 0.046 moles of KClO3 will release (3/2)(0.0462) = 0.0693 moles of O2
1 mole of O2 at STP occupies 22.4 L
So 0.069 moles of O2 will occupy 0.0693 * 22.4 = 1.55L
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