Question

A 550mL sample of oxygen gas is collected over water at a temperature of 21.0°C and a pressure by f 743mmHg according to the following reaction.
(vapor pressure of water at 21.0°C=18.7mmHg, 1atm=760mmHg)

2KCLO3+(s) 2KCL +(s) + 3O2(g)

a. How many moles of oxygen gas was collected?

b. How many grams of potassium chlorate reacted?

Answer 1

(a) We know that ideal gas equation is PV = nRT

Where

T = Temperature = 21 oC = 21 + 273 = 294 K

P = pressure = vapour pressure of solution - vapour pressure of water

= 743 - 18.7 mm Hg

= 724.3 mmHg x ( 1 atm / 760 mm Hg )

= 0.953 atm

n = No . of moles = ?

R = gas constant = 0.0821 L atm / mol - K

V= Volume of the gas = 550 mL = 0.550 L

Plug the values we get n = (PV)/(RT)

= 0.022 moles

So the number of moles of Oxygen collected is 0.022 moles

(b) 2KClO3+(s) ----> 2KCl +(s) + 3O2(g)

From the above balanced reaction,

3 moles of O2 produced along with 2 moles of KCl

0.022 moles of O2 produced along with M moles of KCl

M = ( 0.022 x 2 ) / 3

= 0.015 moles of  potassium chlorate

Mass of  potassium chlorate , m = Number of moles x molar mass

= 0.015 mol x 122.55 g/mol

= 1.80 g