Chapter 9 Calculations from Chemical Equations

Chapter 9
Calculations from
Chemical Equations
1

9.1 A short review
molar mass – the mass of a mole of any formula
unit
relationship between molecule and mole
1 mole = 6.022 × 10 23 molecules
6.022 × 10 23 formula units
6.022 × 10 23 atoms
6.022 × 10 23 ions
grams of a substance
molar mass = ————————————
number of moles of the substance
number of molecules
number of moles = ——————————
6.022 × 10 23 molecules/mole
in a balanced equations, the number in front of a
formula represents the number of moles of that
substance in the reaction
2

9.2 Introduction to stoichiometry
stoichiometry – the chemistry dealing with
quantitative relationships among reactants
and products
mole ratio – a ratio between the number of moles
of any two species involved in a chemical
reaction
ex. 2 H 2 + O 2 2 H 2 O
2 mol H 2 1 mol O 2 2 mol H 2 O
three steps in mole-ratio method:
* write the balanced equation
step 1. determine the number of moles of
starting substance
grams of starting substance
moles = ——————————————
molar mass of starting substance
step 2. determine the mole ratio of the
desired substance to the starting substance
moles of desired substance
in the equation
mole ratio = ————————————
moles of desired substance
in the equation
3

step 3. calculate the desired substance in the
units specified in the problem
moles of desired
substance
moles of starting
substance
moles of desired substance
in the equation
= ————————
moles of starting substance
in the equation
4

9.3 Mole-mole calculations
ex. 9.1 C 6 H 12 O 6 + 6O 2 6CO 2 + 6H 2 O
complete reaction of 2.0 mol of glucose,
how many moles of CO 2 produced
mole ratio: 1 mol C 6 H 12 O 6 6 mol CO 2
6 mol CO 2
2 mol C 6 H 12 O 6 —————— = 12 mol CO 2
1 mol C 6 H 12 O 6
ex. 9.2 3H 2 + N 2 2NH 3
how many moles of NH 3 produced from
8.00 mol of H 2 reacting with N 2
mole ratio: 3 mol H 2 2 mol NH 3
2 mol NH 3
8 mol H 2 ————— = 5.33 mol NH 3
3 mol H 2
ex. 9.3 K 2 Cr 2 O 7 + 6KI + 7H 2 SO 4
Cr 2 (SO 4 ) 3 + 4K 2 SO 4 + 3I 2 + 7H 2 O
how many moles of K 2 Cr 2 O 7 will react
with 2.0 mol of KI
mole ratio: 1 mol K 2 Cr 2 O 7 6 mol KI
1 mol K 2 Cr 2 O 7
2 mol KI ————— = 0.33 mol K 2 Cr 2 O 7
6 mol KI
5

9.4 Mole-mass calculations
ex. 9.5 what mass of H 2 can be produced by
reacting 6.0 mol of Al with HCl
2Al + 6HCl 2AlCl 3 + 3H 2
mole ratio: 2 mol Al 3 mol H 2
3 mol H 2
6 mol Al ————— = 9 mol H 2
2 mol Al
9 mol × 2.016 g/mol = 18 g H 2
ex. 9.6 how many mole of water can be
produced by burning 325 g of octane
2C 8 H 18 + 25O 2 16CO 2 + 18H 2 O
mole ratio: 2 mol C 8 H 18 18 mol H 2 O
325 /114.2 = 2.85 mol C 8 H 18
18 mol H 2 O
2.85 mol C 8 H 18 ————— = 25.6 mol H 2 O
2 mol C 8 H 18
6

9.5 Mass-mass calculations
ex. 9.7 what mass of CO 2 can be produced by
combustion of 100 g of pentane
C 5 H 12 + 8O 2 5CO 2 + 6H 2 O
100 / 72.15 = 1.39 mol C 5 H 12
mole ratio: 1 mol C 5 H 12 5 mol CO 2
5 mol CO 2
1.39 mol C 5 H 12 ————— = 6.95 mol CO 2
1 mol C 5 H 12
6.95 mol × 44.01 g/mol = 305 g H 2
ex. 9.8 4Zn + 10HNO 3
4Zn(NO 3 ) 2 + N 2 O + 5H 2 O
how many grams of HNO 3 are required
to produce 8.75 g of N 2 O
8.75 / 44.02 = 0.199 mol N 2 O
mole ratio: 10 mol HNO 3 1 mol N 2 O
10 mol HNO 3
0.199 mol N 2 O ————— =1.99 mol HNO 3
1 mol N 2 O
1.99 mol × 63.02 g/mol = 125 g HNO 3
7

9.6 Limiting reactant and yield
calculations
limiting reactant – limits the amount of product
that can be formed
ex. H 2 + Cl 2
2HCl
limiting reactant: H 2
8

identify the limiting reactant as following:
1. calculate the amount of product formed from
each reactant
2. determining which reactant is limiting (the
one gives the least amount of product)
3. calculate the amount of the other reactant
required to react with the limiting reactant
calculate the amount of that substance
remains unreacted
ex. 9.9 how many moles of HCl can be
produced by reacting of 4 mol H 2 and
3.5 mol Cl 2
Which compound is limiting reactant
mole ratio: 1 mol H 2 1 mol Cl 2 2 mol HCl
2 mol HCl
4 mol H 2 ————— = 8 mol HCl
1 mol H 2
2 mol HCl
3.5 mol Cl 2 ————— = 7 mol HCl
1 mol Cl 2
7 mol HCl produced
Cl 2 is limiting reactant
9

ex. 9.10 how many moles of Fe 3 O 4 can be
obtained by reacting of 16.8 g Fe with
10 g H 2 O
Which compound is limiting reactant
3Fe + 4H 2 O Fe 3 O 4 + 4H 2
mole ratio:
3 mol Fe 4 mol H 2 O 1 mol Fe 3 O 4
1 mol Fe 3 O 4
(16.8/55.85) mol Fe ———— = 0.1 mol Fe 3 O 4
3 mol Fe
1 mol Fe 3 O 4
(10.0/18.02) mol H 2 O ————— = 0.139 mol
4 mol H 2 O Fe 3 O 4
0.1 mol Fe 3 O 4 obtained
Fe is limiting reactant
ex. 9.11 how many grams of AgBr can be formed
when solutions containing 50 g MgBr 2
and 100 g AgNO 3 are mixed together
how many grams of the excess reactant
remain unreacted
MgBr 2 + 2AgNO 3 2AgBr + Mg(NO 3 ) 2
mole ratio:
1 mol MgBr 2 2 mol AgNO 3 2 mol AgBr
2 mol AgBr
(50/184.1) mol MgBr 2 ———— (187.8) = 102 g AgBr
1 mol MgBr 2
10

2 mol AgBr
(100/169.9) mol AgNO 3 ———— (187.8) = 110.5 g
2 mol AgNO 3 AgBr
102 g AgBr is yielded
MgBr 2 is limiting reactant
2 mol AgNO 3
(50/184.1) mol MgBr 2 ————— (169.9) = 92.3 g
1 mol MgBr 2 AgNO 3
100 -92.3 = 7.7 g AgNO 3 unreacted
theoretical yield – calculated amount of product
according to the chemical equation
actual yield – amount of product experimentally
obtained
actual yield
percent yield = ─────── × 100%
theoretical yield
ex. 9.12 CCl 4 was prepared by reacting 100 g
CS 2 and 100 g Cl 2 , calculate the percent
yield if 65 g CCl 4 obtained
CS 2 + Cl 2 CCl 4 + S 2 Cl 2
1 mol CCl 4
(100/76.15) mol CS 2 ———— (153.8) = 202 g CCl 4
1 mol CS 2
1 mol CCl 4
(100/70.9) mol Cl 2 ———— (153.8) = 72.3 g CCl 4
1 mol Cl 2
11

theoretical yield is 72.3 g CCl 4
percent yield = (65 / 72.3) × 100% = 89.9%
ex. 9.13 AgBr was prepared by reacting 200 g
MgBr 2 and adequate amount of AgNO 3 ,
calculate the percent yield if 375 g AgBr
formed
MgBr 2 + 2AgNO 3 2AgBr + Mg(NO 3 ) 2
2 mol AgBr
(200/184.1) mol MgBr 2 ————— (187.8) = 408 g AgBr
1 mol MgBr 2
percent yield = (375 / 408) × 100% = 91.91%
12