Answer the following stoichiometry questions by referring to the equation below:
2  KClO3  →  2 KCl + 3 O2
i.     If 1.50 mol of KClO3 decomposes, what is the mass of O2 that will be produced?
ii.     If 80.0 grams of O2 was produced, how many moles of KClO3 are decomposed?
iii.    Find the mass of KClO3 needed if we need to produce 2.75  mol of KCl.
This is a problem of determining mole ratios in the stoichiometric equations.
(i) Mass of O2
We have moles of KClO3 = 1.50mol
Mole ratios of KClO3 : O2 = 2:3
Thus moles of O2 "=\\dfrac{3}{2}x1.50mol"
= 2.25mol of O2
Moles "=\\dfrac{Mass}{MM}"
Therefore Mass of O2 = Moles x MM
= 2.25mol x 32g/mol
= 72g
(ii) Moles of KClO3
Moles of O2 "=\\dfrac{mass}{MM} = \\dfrac{80g}{32g\/mol} = 2.5mol"
Mole ratios of O2:KClO3 = 3:2
Moles of KClO3 "=\\dfrac{2}{3}x 2.5mol = 1.67 mol"
(ii)Mass of KClO3
We are given moles of KCl = 2.75mol
Mole ratios of KCl : KClO3 = 2:2
Thus moles of KClO3 "=\\dfrac{2}{2}x2.75mol = 2.75 mol"
Mass = Moles x MM
= 2.75mol x 122.55g/mol
= 337.013g
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