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# Answer to Question #22447 in Organic Chemistry for Ryan

Question #22447
Solid potassium chlorate (KClO3) decomposes to produce solid potassium chloride and oxygen gas. What volume of oxygen gas, measured at 40 C and 655 mm Hg, will be produced when 13.5 of potassium chlorate is decomposed?
1
Expert's answer
2013-01-21T08:24:36-0500
Theequation of the reaction is
2KClO3 = 2KCl + 3O2
If m(KClO3) = 13.5 g so we can find the amount of substance of KClO3
n(KClO3) = m/M
M(KClO3) = 39 + 35.5 + 16*3 = 122.5 g / mol
n(KClO3) = 13.5 g/ 122.5 g/mol = 0.11 mol
The relationship between n(KClO3) and n(O2) is
n(KClO3) : n(O2) = 2 : 3;
n(O2) = n(KClO3) *3/2
n(O2) = 0.11*3/2 = 0.165 moles;
pV = nRT
from this formula
V = nRT/p;
V(O2) = 0.165 * 8.314 * 313 / 87 330 = 0.004916m^3 or 4.916 L

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