Question

Solid potassium chlorate decomposes upon heating to form

Answer 1

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

$\frac{P1V1}{T1}$ = $\frac{P2V2}{T2}$

V2 = $\frac{P1V1T2}{T1P2}$

V2 = 664.1052 ml

245.2   yielded 67.2 molar volume of oxygen

0.66411 will yield = $\frac{245.2 * 0.66411}{67.2}$  = 2.4232 g

percentage of potassium chlorate in the original mixture = $\frac{2.4232 * 100}{7.44}$ = 32.6%