Question

Solid potassium chlorate decomposes upon heating to form

solidpotassium chloride and oxygen gas. A mixture of potassium
chlorateand potassium chloride with a mass of 7.44 g is heated
until allthe potassium chlorate has decomposed. If 789 ml of gas
iscolledcted over water at 0.976 atmospheres and 30.0
C, what is the percent potassium chlorate in theoriginal mixture.

Answer 1

32.6%

Step-by-step 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

`(P1V1)/(T1)` =
`(P2V2)/(T2)`

V2 =
`(P1V1T2)/(T1P2)`

V2 = 664.1052 ml

245.2 yielded 67.2 molar volume of oxygen

0.66411 will yield =
`(245.2 * 0.66411)/(67.2)` = 2.4232 g

percentage of potassium chlorate in the original mixture =
`(2.4232 * 100)/(7.44)` = 32.6%