chapter 10 mass relationships in chemical reactions: stoichiometry
TRANSCRIPT
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%