Answer to Question #22447 in Organic Chemistry for Ryan

Question

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?

Answer 1

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