Answer:
32.6%
Explanation:
Equation of reaction
2KClO₃ (s) → 2KCl (s) + 3O₂ (g)
Molar mass of 2KClO₃ = 245.2 g/mol ( 122.6 × 2)
Molar volume of Oxygen at s.t.p = 22.4L / mol
since the gas was collected over water,
total pressure = pressure of water vapor + pressure of oxygen gas
0.976 = 0.04184211 atm + pressure of oxygen gas at 30°C
pressure of oxygen = 0.976 - 0.04184211 = 0.9341579 atm = P1
P2 = 1 atm, V1 = 789ml, V2 = unknown, T1 = 303K, T2 = 273k at s.t.p
Using ideal gas equation
[tex]\frac{P1V1}{T1}[/tex] = [tex]\frac{P2V2}{T2}[/tex]
V2 = [tex]\frac{P1V1T2}{T1P2}[/tex]
V2 = 664.1052 ml
245.2 yielded 67.2 molar volume of oxygen
0.66411 will yield = [tex]\frac{245.2 * 0.66411}{67.2}[/tex] = 2.4232 g
percentage of potassium chlorate in the original mixture = [tex]\frac{2.4232 * 100}{7.44}[/tex] = 32.6%
