Making Quantitative Predictions About Chemical Reactions Chemical Stoichiometry

Making Quantitative Predictions

About Chemical Reactions

Chemical Stoichiometry

Qualitative and QuantitativePredictions

The classification of chemical reactions helps us to predict what will be produced by a reaction.

We also need to know how much will be produced by a reaction.

It is also necessary to know how much reactant is required to produce a desired amount of product.

The study of the quantitative

Relationships between the reactants and products in a chemical reaction is called

chemical stoichiometry.

Mass-Mass Stoichiometry

The first predictions we will attempt to make will involve the masses of reactants and products.

How many grams of oxygen would be released by the decomposition of 10.00 grams of potassium chlorate?

The Balanced Equation Calculations for chemical reactions must be based upon balanced chemical equations. Write the equation for the decomposition of potassium chlorate to produce oxygen gas and potassium chloride.

KClO3 ==> KCl + O

Remember that oxygen is a diatomic gas.

The Balanced Equation Calculations for chemical reactions must be based upon balanced chemical equations.Write the equation for the decomposition of potassium chlorate to produce oxygen gas and potassium chloride.

KClO3 ==> KCl + O2

That’s better.

The Balanced Equation Calculations for chemical reactions must be based upon balanced chemical equations.Write the equation for the decomposition of potassium chlorate to produce oxygen gas and potassium chloride.

KClO3 ==> KCl + O2

Now balance with coefficients.

The Balanced Equation Every calculations for chemical reactions must be based upon a balanced chemical equation.Write the equation for the decomposition of potassium chlorate to produce oxygen gas and potassium chloride.

2 KClO3 ==> 2 KCl + 3 O2

The Flow Chart

2 KClO3 ==> 2 KCl + 3 O2

Beneath the balanced equation, construct a flow chart to organize the elements of the problem.

After reading the problem a second time, put a question mark under the reactant or product you need to find.

Include the unit of measurement.

The Flow Chart

2 KClO3 ==> 2 KCl + 3 O2

? g

The Flow Chart

2 KClO3 ==> 2 KCl + 3 O2

? g

After the third reading, put the measurements or quantities that you were given under the reactant or product to which that value pertains.

The Flow Chart

2 KClO3 ==> 2 KCl + 3 O2

? g

After the third reading, put the measurements or quantities that you were given under the reactant or product to which that value pertains.

10.00 g

The Flow Chart

2 KClO3 ==> 2 KCl + 3 O2

? g

Because the balanced equation only gives us molar relationships but we are interested in masses, we must convert what the equation gives us into what we want.

10.00 g

The Flow Chart

2 KClO3 ==> 2 KCl + 3 O2

? g10.00 g

Grams of KClO3 must be converted to moles of KClO3.

Moles of KClO3 then convert to moles of O2 .

Moles of O2 then convert to grams of O2 .

Use lines beneath the equation to show these three steps.

The Flow Chart

2 KClO3 ==> 2 KCl + 3 O2

? g10.00 g

Use lines beneath the equation to show these three steps.

||

mol KClO3 ----------- mol O2

||

The Flow Chart

2 KClO3 ==> 2 KCl + 3 O2

? g10.00 g

Each line represents a conversion factor.Arrows show the sequence of conversion factors.

||

\/

mol KClO3 -----------> mol O2

/\||

12

3

The Calculation Process

2 KClO3 ==> 2 KCl + 3 O2

? g10.00 g

Now we set up three conversion factors. Each conversion factor will accomplish one leg of the flow chart.

||

\/

mol KClO3 -----------> mol O2

/\||

12

3

The Calculation Process 2 KClO3 ==> 2 KCl + 3 O2

? g10.00 g\/

mol KClO3 -----------> mol O2

/\|1 2 3|

1) 10.00 g KClO3 mol KClO3 =x ____________122.6

mol KClO3

g KClO3

10.0816


? g10.00 g\/

mol KClO3 -----------> mol O2

/\|1 2 3|

1) 10.00 g KClO3 mol KClO3 =x ____________122.6

mol KClO3

g KClO3

10.0816

The second conversion factor will convert moles of potassium chlorate to moles of oxygen gas.


? g10.00 g\/

mol KClO3 -----------> mol O2

/\|1 2 3|

1) 10.00 mol KClO3 =x ____________122.6

mol KClO3

g KClO3

10.0816 g KClO3

2) 0.0816 mol KClO3

x mol O2__________ =molO2 mol KClO3 3

20.122


? g10.00 g\/

mol KClO3 -----------> mol O2

/\|1 2 3|

1) 10.00 mol KClO3 =x ____________122.6

mol KClO3

g KClO3

10.0816 g KClO3

2) 0.0816 mol KClO3


20.122

The third and final conversion factor converts moles of oxygen to grams of oxygen.


? g10.00 g\/

mol KClO3 -----------> mol O2

/\|1 2 3|

1) 10.00 mol KClO3 =x ____________122.6

mol KClO3

g KClO3

10.0816 g KClO3

2) 0.0816 mol KClO3


20.122

3) 0.122 mol O2 = g O2 x __________g O2

mol O2

32.0

1.03.915

2 KClO3 ==> 2 KCl + 3 O2

? g10.00 g\/

mol KClO3 -----------> mol O2

/\|1 2 3|

1) 10.00 mol KClO3 =x ____________122.6

mol KClO3

g KClO3

10.0816 g KClO3

2) 0.0816 mol KClO3


20.122

3) 0.122 mol O2 = g O2 x __________g O2

mol O2

32.0

1.03.915

This calculation allows us to predict that the decomposition of 10.00 grams of potassium chlorate will produce 1.96 grams of oxygen gas.

1) 10.00 mol KClO3 =x ____________122.6

mol KClO3

g KClO3

10.0816 g KClO3

2) 0.0816 mol KClO3


20.122

3) 0.122 mol O2 = g O2 x __________g O2

mol O2

32.0

1.03.915

To avoid rounding off three times, it is often more convenient for calculator use to arrange our three conversion factors in a linear sequence.

So, Instead of three separate calculations, like this,

it would look more like

10.00 x ____________

122.6

mol KClO3

g KClO3

1 g KClO3

mol KClO3

x __________molO23

2

= g O2

x __________g O2

mol O2

32.0

1.0

3.915

This.

Once set up in this format, the series of calculator keystrokes used is quite simple.

10.00 x ____________

122.6

mol KClO3

g KClO3

1 g KClO3

mol KClO3 x __________

molO23

2

= g O2

x __________g O2

mol O2

32.0

1.0

3.915

Remember to multiply by numerators (the numbers on the top) and divide by denominators (the numbers on the bottom).

10.00 x ____________

122.6

mol KClO3

g KClOg KClO33

1 g KClOg KClO33

mol KClO3 x __________

molO23

2

= g O2

x __________g O2

mol O2

32.0

1.0

3.915


10.00 x ____________

122.6

mol KClOmol KClO33

g KClOg KClO33

1 g KClOg KClO33

mol KClOmol KClO33 x __________

molO23

2

= g O2

x __________g O2

mol O2

32.0

1.0

3.915


10.00 x ____________

122.6

mol KClOmol KClO33

g KClOg KClO33

1 g KClOg KClO33

mol KClOmol KClO33 x __________

molOmolO223

2

= g O2

x __________g O2

mol Omol O22

32.0

1.0

3.915


10.00 x ____________

122.6

mol KClO3

g KClO3

1 g KClO3

mol KClO3

x __________molO23

2

= g O2

x __________g O2

mol O2

32.0

1.0

3.915

10 / 122.6 x 3 / 2 x 32 = 3.91517289

Rounding to four significant figures (because the mass data was given that way) gives us

3.915 g O2

Determine the mass of iron which

could be produced by the reaction of 100.0 grams of iron(III) oxide with enough carbon to use it all.

Fe2O3 + C ==> Fe + CO2



2 Fe2O3 + 3 C ==> 4 Fe + 3 CO2



2 Fe2O3 + 3 C ==> 4 Fe + 3 CO2

? g



2 Fe2O3 + 3 C ==> 4 Fe + 3 CO2

? g100.0 g



2 Fe2O3 + 3 C ==> 4 Fe + 3 CO2

? g100.0 g

mol Fe2O 3 ------------------------> mol Fe

||\/

||

/\



2 Fe2O3 + 3 C ==> 4 Fe + 3 CO2

? g100.0 g

mol Fe2O 3 ------------------------> mol Fe

||\/

||

/\1

23

100.0 g Fe2O 3 x _________ x ________ x ________ 1 mol Fe2O 3

g Fe2O 3

mol Fe

mol Fe2O 3

g Fe

1 mol Fe159.6

4

2

55.8

100 / 159.6 x 2 x 55.8 = 69.9 g Fe

100.0 g FeFe22OO 3 3 x _________ x ________ x ________ 1 mol Fe2O 3

g Feg Fe22OO 3 3

mol Fe

mol Fe2O 3

g Fe

1 mol Fe159.6

4

2

55.8

100 / 159.6 x 2 x 55.8 = 69.9 g Fe

100.0 g FeFe22OO 3 3 x _________ x ________ x ________ 1 mol Femol Fe22OO 3 3

g Feg Fe22OO 3 3

mol Fe

mol Femol Fe22OO 3 3

g Fe

1 mol Fe159.6

4

2

55.8

100 / 159.6 x 2 x 55.8 = 69.9 g Fe

100.0 g FeFe22OO 3 3 x _________ x ________ x ________ 1 mol Femol Fe22OO 3 3

g Feg Fe22OO 3 3

mol Femol Fe

mol Femol Fe22OO 3 3

g Fe

1 mol Femol Fe159.6

4

2

55.8

100 / 159.6 x 2 x 55.8 = 69.9 g Fe

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Making Quantitative Predictions About Chemical Reactions Chemical Stoichiometry