Making Quantitative Predictions About Chemical Reactions Chemical Stoichiometry

Qualitative and Quantitative Predictions 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 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. KClO 3 ==> 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. KClO 3 ==> KCl + O 2 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. KClO 3 ==> KCl + O 2 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 KClO 3 ==> 2 KCl + 3 O 2

The Flow Chart 2 KClO 3 ==> 2 KCl + 3 O 2 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 KClO 3 ==> 2 KCl + 3 O 2 ? g

The Flow Chart 2 KClO 3 ==> 2 KCl + 3 O 2 ? 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 KClO 3 ==> 2 KCl + 3 O 2 ? g After the third reading, put the measurements or quantities that you were given under the reactant or product to which that value pertains g

The Flow Chart 2 KClO 3 ==> 2 KCl + 3 O 2 ? 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 g

The Flow Chart 2 KClO 3 ==> 2 KCl + 3 O 2 ? g g Grams of KClO 3 must be converted to moles of KClO 3. Moles of KClO 3 then convert to moles of O 2. Moles of O 2 then convert to grams of O 2. Use lines beneath the equation to show these three steps.

The Flow Chart 2 KClO 3 ==> 2 KCl + 3 O 2 ? g g Use lines beneath the equation to show these three steps. | | mol KClO mol O 2 | |

The Flow Chart 2 KClO 3 ==> 2 KCl + 3 O 2 ? g g Each line represents a conversion factor. Arrows show the sequence of conversion factors. |||| \/ mol KClO > mol O 2 /\ |||| 1 2 3

The Calculation Process 2 KClO 3 ==> 2 KCl + 3 O 2 ? g g Now we set up three conversion factors. Each conversion factor will accomplish one leg of the flow chart. |||| \/ mol KClO > mol O 2 /\ |||| 1 2 3

The Calculation Process 2 KClO 3 ==> 2 KCl + 3 O 2 ? g g \/ mol KClO > mol O 2 /\ | 12 3 | 1) g KClO 3 mol KClO 3 =x ____________ mol KClO 3 g KClO

The Calculation Process 2 KClO 3 ==> 2 KCl + 3 O 2 ? g g \/ mol KClO > mol O 2 /\ | 12 3 | 1) g KClO 3 mol KClO 3 =x ____________ mol KClO 3 g KClO The second conversion factor will convert moles of potassium chlorate to moles of oxygen gas.

The Calculation Process 2 KClO 3 ==> 2 KCl + 3 O 2 ? g g \/ mol KClO > mol O 2 /\ | 12 3 | 1) mol KClO 3 =x ____________ mol KClO 3 g KClO g KClO 3 2) mol KClO 3 x mol O 2 __________= molO 2 mol KClO

The Calculation Process 2 KClO 3 ==> 2 KCl + 3 O 2 ? g g \/ mol KClO > mol O 2 /\ | 12 3 | 1) mol KClO 3 =x ____________ mol KClO 3 g KClO g KClO 3 2) mol KClO 3 x mol O 2 __________= molO 2 mol KClO The third and final conversion factor converts moles of oxygen to grams of oxygen.

The Calculation Process 2 KClO 3 ==> 2 KCl + 3 O 2 ? g g \/ mol KClO > mol O 2 /\ | 12 3 | 1) mol KClO 3 =x ____________ mol KClO 3 g KClO g KClO 3 2) mol KClO 3 x mol O 2 __________= molO 2 mol KClO )0.122 mol O 2 = g O 2 x__________ g O 2 mol O

2 KClO 3 ==> 2 KCl + 3 O 2 ? g g \/ mol KClO > mol O 2 /\ | 12 3 | 1) mol KClO 3 =x ____________ mol KClO 3 g KClO g KClO 3 2) mol KClO 3 x mol O 2 __________= molO 2 mol KClO )0.122 mol O 2 = g O 2 x__________ g O 2 mol O This calculation allows us to predict that the decomposition of grams of potassium chlorate will produce 1.96 grams of oxygen gas.

1) mol KClO 3 =x ____________ mol KClO 3 g KClO g KClO 3 2) mol KClO 3 x mol O 2 __________= molO 2 mol KClO )0.122 mol O 2 = g O 2 x__________ g O 2 mol O 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

10.00x ____________ mol KClO 3 g KClO 3 1 mol KClO 3 x __________ molO = g O 2 x __________ g O 2 mol O This. Once set up in this format, the series of calculator keystrokes used is quite simple.

10.00 x ____________ mol KClO 3 g KClO 3 1 mol KClO 3 x __________ molO = g O 2 x __________ g O 2 mol O Remember to multiply by numerators (the numbers on the top) and divide by denominators (the numbers on the bottom).

10.00 x ____________ mol KClO 3 1 x __________ molO = g O 2 x __________ g O 2 mol O Remember to multiply by numerators (the numbers on the top) and divide by denominators (the numbers on the bottom).

10.00 x ____________ x __________ molO = g O 2 x __________ g O 2 mol O Remember to multiply by numerators (the numbers on the top) and divide by denominators (the numbers on the bottom).

10.00 x ____________ x __________ 3 2 = g O 2 x __________ g O Remember to multiply by numerators (the numbers on the top) and divide by denominators (the numbers on the bottom).

10.00x ____________ mol KClO 3 g KClO 3 1 mol KClO 3 x __________ molO = g O 2 x __________ g O 2 mol O / x 3 / 2 x 32 = Rounding to four significant figures (because the mass data was given that way) gives us 3.915g O 2

Determine the mass of iron which could be produced by the reaction of grams of iron(III) oxide with enough carbon to use it all. Fe 2 O 3 + C ==> Fe + CO 2

Determine the mass of iron which could be produced by the reaction of grams of iron(III) oxide with enough carbon to use it all. 2 Fe 2 O C ==> 4 Fe + 3 CO 2

Determine the mass of iron which could be produced by the reaction of grams of iron(III) oxide with enough carbon to use it all. 2 Fe 2 O C ==> 4 Fe + 3 CO 2 ? g

Determine the mass of iron which could be produced by the reaction of grams of iron(III) oxide with enough carbon to use it all. 2 Fe 2 O C ==> 4 Fe + 3 CO 2 ? g g

Determine the mass of iron which could be produced by the reaction of grams of iron(III) oxide with enough carbon to use it all. 2 Fe 2 O C ==> 4 Fe + 3 CO 2 ? g g mol Fe 2 O > mol Fe |||| \/ |||| /\

Determine the mass of iron which could be produced by the reaction of grams of iron(III) oxide with enough carbon to use it all. 2 Fe 2 O C ==> 4 Fe + 3 CO 2 ? g g mol Fe 2 O > mol Fe |||| \/ |||| /\ 1 2 3

100.0 g Fe 2 O 3 x _________ x ________ x ________ 1 mol Fe 2 O 3 g Fe 2 O 3 mol Fe mol Fe 2 O 3 g Fe 1 mol Fe / x 2 x 55.8 =69.9g Fe

1 mol Fe 2 O 3 mol Fe mol Fe 2 O 3 g Fe 1 mol Fe / x 2 x 55.8 =69.9g Fe

mol Fe g Fe 1 mol Fe / x 2 x 55.8 =69.9g Fe

/ x 2 x 55.8 =69.9g Fe