Gas Laws: Stoichiometry

Question

1. Potassium chlorate, when heated, decomposes to form potassium chloride and oxygen gas:

___KClO3(s) ---> ___O2(g) + ___KCl(s)

a.) Balance the equation.

b.) How many grams of KClO3 are needed to produce 0.10027 g of KCl?

c.) What was the volume of the oxygen gas (in mL) produced in part b? The gas is collected over water with a total pressure of 1.152 atm at 23.10 degrees Celsius. The vapor pressure of water is 21.12 torr at 23.10 degrees Celsius.

Answer 1

a) 2KClO₃(s) ==> 3O₂(g) + 2KCl(s) ... balanced equation

b) Using dimensional analysis, we can find mass of KClO3 needed. The following analysis converts g KCl to moles of KCl; then using the stoichiometry of the balanced equation, converts moles KCl to moles of KClO₃ and finally to grams of KClO₃.

0.10027 g KCl x 1 mol KCl/74.55 g x 2 mol KClO₃/2 mol KCl x 122.55 g/mol KClO₃ = 0.16483 g KClO₃

c) First, we find the moles of O₂ formed using the stoichiometry of the balanced equation knowing that 3 moles of O2 are formed from 2 moles of KCl. We then use the ideal gas law, PV = nRT to find the volume.

From part (b) we have 0.10027 g KCl x 1 mol KCl/74.55 g = 0.0013450 moles KCl

Moles O₂ formed = 0.0013450 moles KCl x 3 moles O₂/2 moles KCl = 0.0020175 moles O₂

PV = nRT

P = 1.152 atm - (21.12 torr x 1 atm/760 torr) = 1.152 atm - 0.03 atm = 1.12 atm (NOTE: convert torr to atm and subtract the vapor pressure of water from the total pressure to get the pressure of O₂)

V = ?

n = 0.0020175 moles

R = 0.0821 Latm/Kmol

T = 23.10ºC + 273.15 = 296.25K

Solving for V, we have...

V = nRT/P = (0.0020175)(0.0821)(296.25)/1.12

V = 0.04907 L = 49.1 mls