Transcript Ionic Reactions in Aqueous Solutions

Exp 9: Analysis of a KClO3-KCl Mixture
• Introduction
– Small amounts of O2 can be generated from decomposition
reaction
D
• 2BaO2(s) 2BaO(s) + O2(g)
D
• 2KClO3(s)  2KCl(s) + 3O2(g)
MnO2
Manganese oxide (MnO2) is a catalyst
Objectives
• Study thermal decomposition of potassium chlorate,
KClO3
• Identify product of reaction from the changes in mass of
the reactant
• Perform calculations based on balanced equations
Exp 9: Analysis of a KClO3-KCl Mixture
• Balancing equations
2 Al(s) + 6 HCl(aq)  2 AlCl3 (aq) + 3 H2(g)
2 mol Al(s) + 6 mol HCl(aq)  2 mol AlCl3 (aq) + 3 mol H2(g)
• Balanced reaction equation is a mole statement
 2 mol Al is consumed to produce 3 mol H2 (2:3 ratio)
 6 mol HCl is consumed to produce 2 mol AlCl3 (ratio 3:1) and 3 mol H2
(ratio 2:1)
 if 1 mol Al is consumed then only 1.5 mol H2 is produced (2:3 ratio)
• Molar ratios 2 mol Al / 6 mol HCl
6 mol HCl / 3 mol H2
2 mol AlCl3 / 3 mol H2
6 mol HCl / 2 mol AlCl3
Exp 9: Analysis of a KClO3-KCl Mixture
• Remember
– If you need quantities in gram: convert mole to
grams by multiplying by molar mass
– If you need mol: convert gram to mole by
dividing by the molar mass
Exp 9: Analysis of a KClO3-KCl Mixture
• Calculations with balanced equations, use molar ratios
2 Al(s) + 6 HCl(aq)  2 AlCl3 (aq) + 3 H2(g)
2 mol Al  2 mol AlCl3 + 3 mol H2
1 mol Al  1 mol AlCl3 + 1.5 mol H2
How many mole of Al to produce 5 mol H2?
 2 mol Al/3 mol H2 * 5 mol H2 = 3.333 mol Al
Exp 9: Analysis of a KClO3-KCl Mixture
Calculations with balanced equations
• What mass of Al to produce 5.00 g H2?
• Step 1) convert mass H2 to mol H2
• Step 2) Calculate amount of Al in mol based on molar ratio
• Step 3) Calculate amount of Al in g, based on molar mass of Al
 5.00 g H2 = 5.00 g / 2.00 g/mol H2 = 2.50 mol H2
 Molar ration Al/H2 = 2 mol Al per 3 mol H2
 2.50 mol H2 = 2.50 mol * 2/3 * 1 mol Al = 1.67 mol Al
Molar mass of Al = 27.3 g/mol  mass of Al in g = 1.67 mol * 27.3
g/mol = 45.50 g Al
Exp 9: Analysis of a KClO3-KCl Mixture
How much AlCl3 is produced?
 2 mol Al produces 2 mol AlCl3
 molar mass of AlCl3 = molar mass of Al + 3 x molar mass
of Cl = 27.3 g/mol + 3 x 35.45 g/mol = 133.65 g/mol AlCl3
 1.67 mol Al * (2 mol AlCl3/2 mol Al) * 133.65 g/mol =
223.20 g AlCl3
Exp 9: Analysis of a KClO3-KCl Mixture
Procedure:
• Determine the amount of KClO3 in a mixture of KClO3 and KCl by
determining the change in mass when O2 is released
• Calculate the molar mass based on the molar ratio of KClO3 and O2 in
the balanced equation
D
• 2KClO3(s)  2KCl(s) + 3O2(g)
D
MnO2
• KCl(s)  no reaction (there is no O2 to evolve!)
MnO2
 The amount of KClO3 decreases, whereas the amount of KCl
increases
 The amount of KClO3 decomposed is proportional to the amount
of O2 generated
 The amount of O2 generated is proportional to the amount of
weight loss of the whole sample
Exp 9: Analysis of a KClO3-KCl Mixture
Procedure:
– Put a small amount (1-2 g) of MnO2 (catalyst) in a test
tube and heat over a hot flame for ~ 5 min to drive out
moisture and organic components (explosion risk)
– Cool test tube + MnO2 to room temperature
– Weigh test tube on a balance, and record mass
– Add KClO3 + KCl mixture to tube, mix thoroughly and
weigh tube again
Exp 9: Analysis of a KClO3-KCl Mixture
Experimental
– Clamp tube in a clamp in a ringstand under
45o angle
– Heat contents with a moderate flame
– If white smoke evaporates, remove flame
• This is KCl that evaporates and escapes
and it will affect your results!
– After about 8 minutes, heat the mixture with
a hot flame for about 10 more minutes
– Allow the tube to cool to room temperature
– Weigh the test tube with the mixture
Exp 9: Analysis of a KClO3-KCl Mixture
Procedure:
– Reheat for an additional 10 minutes
and weigh again to make sure that all
O2 was driven out
– Two successive heatings need to
result in a difference of 0.05 g in
weight or less
– Use the results of the last weighing to
calculate the mass of O2 and the mass
of KClO3
Exp 9: Analysis of a KClO3-KCl Mixture
Calculations:
2KClO3(s)  2KCl(s) + 3O2(g)
Mass of O2 evolved =
Molar ratio of KClO3 : O2 = 2:3
Moles of KClO3 =
Mass of KCl = Total mass – gram KClO3 =
=
mol O2
=
mol KClO3
=
grams KClO3
=
grams KCl
%KClO3 = mass KClO3/total mass (KClO3 + KCl) *100%
=
% KClO3
=
%KCl
%KCl = mass KCl/total mass (KClO3 + KCl) * 100%
For next class
•
•
•
•
Results and answers due next week 04/9/13
Format: See ACES or class website
Include report sheet and calculations
Use your critical thinking skills to discuss
errors
• Pre-lab next week: Absorption Spectrum