49 grams of potassium chlorate is heated calculate the amount of solid and gaseous product formed
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2KClO3 ->2 KCl + 3O2
Moles of KClO3 =
Given mass/ molar mass
49/(31+35.5+48)
49/114.5 = 0.43 moles
moles of kclo3 = moles of kcl formed
= 0.43
Weight of kcl -> 0.43*66.5=28.5g
Weight of O2= 1.5 * 0.43 * 32
[1.5 because 2 reacts with 3, 1 will react with 1.5]
=20.64g
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