Question

Oxygen gas can be prepared by heating potassium chlorate: 2KClO3(s)2KCl(s) + 3O2(g) In one experiment, a sample of KClO3 reacts and the gas produced is collected by water displacement. The gas sample has a temperature of 24.00 °C, a volume of 564.0 mL, and a pressure of 754.0 mm Hg.
Calculate the amount (in moles) of oxygen gas produced in the reaction. The vapor pressure of water is 22.38 mm Hg at 24.00 °C.

Answer 1

Answer:

0.0222 mol

Explanation:

When a gas is collected by water displacement, the total pressure is equal to the pressure of the gas (in this case oxygen) and the pressure of the water vapor.

P = Pw + PO₂

PO₂ = P - Pw = 754.0 mmHg - 22.38 mmHg = 731.6 mmHg

We can find the moles of oxygen using the ideal gas equation.

$P.V=n.R.T\\n=\frac{P.V}{R.T} =\frac{731.6mmHg\times 0.5640L }{(0.08206atm.L/mol.K)\times (24.00 + 273.15)K } \times \frac{1atm}{760mmHg}=0.0222mol$