Oxygen gas can be prepared by heating potassium chlorate acc

Oxygen gas can be prepared by heating potassium chlorate according to the following equation: 2KCIOs(s)2KC(s)+30 ( The product gas, O, is collected over water at a temperature of 25 °C and a pressure of 755 mm Hg. If the wet Os gas formed occupies a volume of 5.38 L, the number of moles of KCIO, reacted was mol The vapor pressure of water is 23.8 mm Hg at 25 °C Submit Answer Retry Entire Group 9 more group attempts remaining

Solution

ANSWER:

Q1..

CONCEPT: Calculate the no. of moles of oxygen , then count the number of moles of potassuim chlorate from stiochiometry.

SOLUTION: The pressure of dry O₂ gas = total pressure - Vapour pressure of water = 755mmHg - 23.8 mmHg = 731.2 mmHg

Convert mmHg to Pa as, 1 mmHg = 133.3 Pa

Therefore 731.2 mmHg = 731.2 X 133.3 = 97468.96 Pa

Volume of O₂ = 5.38L = 5.38 X 10^-3 m³.

Using equation PV = nRT

n = PV / RT = 97468.96 Pa X 5.38 X 10^-3 m³ / 8.314 J/Kmol X 298K = 0.21 moles.

As per balancd chemical equation given in the question 3 moles of O₂ are produced by 2 moles of KClO₃.

3 moles of O₂ = 2 moles of KClO₃.

1 moles of O₂ = 2 / 3 moles of KClO₃.

So 0.211 moles of O₂ = (2 / 3 X 0.211) moles of KClO₃. = 0.14 moles of KClO₃

The answer is 0.14 moles of KClO₃ .

Q2.

Moles of H₂ produced = PV / RT

P = 752 - 23.8 = 728.2 mmHg = 97069.06 Pa

V = 9.63 X 10^-3 m³

moles of H₂ = = 97069.06 Pa X 9.63 X 10^-3 m³ / 8.314 J/molK X 298K = 0.377 moles

Mass of H2 = no. of moles X molar mass = 0.377 X 2g = 754g.