Chapter 10

Mass relationships in chemical reactions:

Stoichiometry

To Review How many grams in a mole of

CaSO4

Al2(SO4)3

Ca 1 x 40.1 g/mol = 40.1 g

S 1 x 32.1 = 32.1

O 4 x 16.0 = 64.0

136.2 g/mol

Al 2 x 27.0 g/mol = 24.0

S 3 x 32.1 = 96.3

O 12 x 16.0 = 192.0

312.3 g/mol

2Al(s) + Al(s) + 33BrBr22(l) (l) 11AlAl22BrBr66(s)(s) Stoichiometric Coefficients- the coefficients

(number in front of the chemical formula) in a balanced chemical equation

Stoichiometry- the relationship between the quantities of chemical reactants and products

Stoichiometric factor- a mole ratio relating moles of one substance to another substance involved in the same chemical reaction

Create two stoichiometric factors that relate the amount of the reactant H2O2 to the product O2

22 H H22OO22(liq) ---> 2 H(liq) ---> 2 H22O(l) + O(l) + 11 OO22(g)(g)

22 mol H mol H22OO2 2 OR OR 11 mol O mol O22

11 mol O mol O22 22 molH molH22OO2 2

Create two stoichiometric factors that relate the amount of the product H2O to the reactant H2O2

22 H H22OO22(l) ---> (l) ---> 22 H H22O(l) + OO(l) + O22(g)(g)

22 mol H mol H22OO2 2 OR OR 22 mol H mol H22OO

22 mol H mol H22OO 22 mol H mol H22OO22

Create two stoichiometric factors that relate the amount of the product CO2 to the reactant C4H10

2 C4H10 + 13O2 -----> 10H2O + 8CO2

88 mol CO mol CO2 2 OR OR 22 mol mol C4H10

22 mol mol C4H10 88 mol CO mol CO2 2

Stoichiometric Calculations The most common types of equation

stoichiometry problems are called 3 step problems Given the mass of a reactant or product determine

the equivalent mass of a different reactant or product

These are called mass to mass or gram to gram conversions

Consider: 2 H2O -------> 2 H2 + O2

How many g of H2 are produced by decomposing 1.00 g H2O?

Stoichiometric Calculations

Mass H2O Mol H2O

Mol H2 Mass H2

1 mol

MM in g

MM in g

1 mol

mol H2

mol H2O

All calculations are done using conversion factors

•Each arrow represents a conversion factor

•Align units first, then place numbers into the

conversion factor

Stoichiometric Calculations

Step 1 - Convert g of given reactant or product to moles Use molar mass of the given

Step 2 - Convert from moles of the given to moles of the unknown Use stoich factor as conversion factor

Step 3 - Convert from moles to g of the unknown Use molar mass of the unknown as conversion

factor

Problem 1a.

If 37.6 g of water is decomposed to hydrogen and oxygen, how many grams of hydrogen will be produced?

Write the balanced chemical equation

2 H2O -------> 2 H2 + O2

Write the information that is given and what's asked for above the equation

Mass 37.6 g ? gEquation 2H2O --------> 2H2 +

O2

Write the molar mass of the substances.

Mass 37.6 g ? gEquation 2 H2O --------> 2H2 +

O2

Molar Masses 18.0 2.0 (g/mol)Remember the Pathway:g H2O mol H2O mol H2 g H2

Step 1 - Convert the given mass of water to moles of water

1 mol H2O = 18.0 g H2O

37.6 g H2O x 1 mol H2O

18.0 g H2O

2 H2O -------> 2 H2 + O2

Step 2 - Convert moles of water to moles of hydrogen using stoichiometric factor

2 H2O -------> 2 H2 + O2

37.6 g H2O x 1 mol H2O x 2 mol H2

18.0 g H2O 2 mol H2O

Step 3 - Convert moles of hydrogen to mass of hydrogen

1 mol H2 = 2.02 g H2

37.6g H2O x 1 mol H2O x 2 mol H2 x 2.0g H2 = 4.17g H2

18.0g H2O 2 mol H2O 1 mol H2

2 H2O -------> 2 H2 + O2

Problem 1b. At room temperature and pressure, aluminum reacts with oxygen to give aluminum oxide. If you react 161 g of Al, what mass of O2 is needed for complete reaction?

Write the chemical equation:

Al(s) + O2(g) -----> Al2O3

2

You need a balanced chemical equation:

Al(s) + O2(g) -----> Al2O324 3

Mass 161 g ? gEquation 4Al(s) + 3O2(g) ---->2Al2O3

Molar Masses 27.0 32.0 (g/mole)

Record given mass and what is asked for in the problem

3 Step Calculations

161 g Al

x 1 mol Al 27.0 g Al

x 3 mol O2

4 mol Al

x 32.0 g O2

1 mol O2

4Al(s) + 3O2(g) ---->2Al2O3

= 143 g O2

161 g Al ? g O2

27.0 g/mol 32.0 g/mol

g Al mol Al mol O2 g O2

What mass of MgO can be produced from igniting 1.5 g of Mg in oxygen?

Mass 1.5 g ? gEquation 2 Mg(s) + O2(g) ---> 2

MgOMolar Masses 24.3 40.3 (g/mole)

1.5 g Mg x 1 mol Mg x24.3 g Mg

2 mol MgO x 2 mol Mg

40.3 g MgO1 mol MgO

= 2.49 g MgO

Theoretical vs Actual Yield Theoretical yield = the amount of

product that could be produced if the reaction is perfectThe amount that results from the 3 step

processThis never happensYou usually get less than the theoretical

yield Actual yield = The amount of product

you recover when you run the experiment

Limiting and Excess Reactant

In chem reactions, reactants are never present in the exact ratio to ensure they are completely used up in the reaction

You usually run out of one reactant, and the other is in excess (left over)Limiting reactant = the one completely

used upExcess reactant = the one with some left

over

A Chemical Reaction

Reactants Products

All of is used up = Limiting ReactantSome of is left over = Excess Reactant

Limiting Reactant Problem Determine the limiting and excess reactant

if 1.40 g of N2 reacts with 1.00 g of H2 to produce ammonia, NH3.

N2 + 3H2 2NH3

1.40 g 1.00 g

28.O g/mol 2.0 g/mol

1.40 g N2 x 3 mol H2

1 mol N2

x 2.0 g H2

1 mol H2

x 1 mol N2

28.0 g N2

= 0.30 g H2

There is more than 0.30 g of H2 present, therefore H2 will be left over (is in excess) and N2 will be thelimiting reactant

Percent Yield

A means of quantifying how efficient a chemical reaction is

% Yield = Actual Yield Theoretical Yield

x 100

% Yields can be above 100%

What’s better, a % yield of 90% or 110% ?

% Yields are usually below 100%less product is recover than is possible

Solving Percent Yield Probs 1 – Determine the limiting reactant 2 – Calculate the theoretical yield from

the given amount of the limiting reactant

3 – Calculate the % yield (the actual yield needs to be given to you in the problem).

% Yield = Actual Yield Theoretical Yield

x 100

Percent Yield Practice Prob #1

6.00 g N2 x 3 mol H2

2 mol N2

x 2.0 g H2

1 mol H2

x 1 mol N2

28.0 g N2

= 0.642 g H2

A student reacts 6.00 g of N2 reacts with 0.500 g of H2

to produce ammonia, NH3. if 2.59 g of NH3 is recovered

from the reaction, what is the student’s % yield?

N2 + 3H2 2NH3

Therefore, 0.500 g H2 is the limiting reactant

Step 1 Determine the limiting reactant.

Percent Yield Practice Prob #1

0.500 g H2 x 2 mol NH3

3 mol H2

x 17.0 g NH3

1 mol NH3

x 1 mol H2

2.0 g H2

= 2.83 g NH3

Step 2 Deter the theoretical yield from limiting reactant

Actual yield = 2.59 g NH3 (given in the problem)

Theoretical yield = 2.83 g NH3

% yield = actual yield x 100 theroretical yield

= 2.59 g x 100 = 91.5% 2.83 g

Lab ExerciseA student reacts 2.40 g of NaHCO3 with an excess of HCLaccording to the reaction:

2.40 g X g

NaHCO3 + HCl ---> NaCl + H2O + CO2

84.0 g/mol 58.5 g/mol

Determine the theoretical yield of NaCl (3 step process)

g NaHCO3 mol NaHCO3 mol NaCl g NaCl

x 1 mol NaHCO3

84.0 g NaHCO3

x 1 mol NaCl

1 mol NaHCO3

x 58.5 g NaCl =

1 mol NaCl

2.40 g NaHCO3

= 1.67 g NaCl

Lab Exercise The reaction from your lab is conducted with a

2.40 g sample of NaHCO3. In the lab, the reaction produces 1.57 g of NaCl. What is the percent yield?

x 1 mol NaHCO3

84.0 g/mol

x 1 mol NaCl

1 mol NaHCO3

x 58.5 g NaCl =

1 mol NaCl

2.40 g NaHCO31.67 g

NaCl

% Yield = actual yield x 100

theoretical yield

= 1.57 g x 100

1.67 g

= 94.0%