CHAPTER 1 - MATTER

from equation:
1 mol C3H8  5 mol O2

mol of O2 = 2.27 x 10-2mol C3H8x  5 mol O2 
 1 mol C3H8 
 

= 0.114 mol O2

mass of O2 = 0.114 mol x 32.0 g mol-1

=3.64 g

150

Example 3

What volume of 0.812 M HCl, in mililiters, is
required to titrate 1.45 g of NaOH to the
equivalence point?

Note:

the pH at which the number of moles of OH-
ions added to a solution is equal
stoichiometrically to the number of moles of
H+ ions originally present.

151

Solution:

HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l)
1.45 g
0.812 M

V= ?

mole of NaOH = 1.45 g
40.0 g mol-1

= 0.0363mol

152

from equation:
1 mole of HCl  1 mole of NaOH

 mole of HCl= mole of NaOH
= 0.0363 mol

volume of HCl = mole of HCl
molarity of HCl

= 0.0363 mol
0.812 mol L-1

= 0.0447 L = 44.7 mL

153

Exercise 1

A 16.50 mL 0.1327 M KMnO4 solution is
needed to oxidize 20.00mL of a FeSO4
solution in an acidic medium. What is the
concentration of the FeSO4 solution? The
net ionic equation is:
5Fe2+ + MnO4- + 8H+  Mn2+ + 5Fe3+ + 4H2O
Answer : 0.5474 M

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Solution

1. check/ write a balance equation.

5Fe 2+ + MnO4- + 8H+  Mn 2+ + 5Fe 3+ + 4H2O

20.00 mL 16.50 mL
?(M) 0.1327M

2. calculate the number of mol

mol MnO   0.1327 x16.5
4 1000

 2.1896 x103 mol

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3. use the equation (mol ratio) to determine the
mole of unknown.

1mol MnO   5 mol Fe 2
4

2.1896 x 10 3 mol MnO   2.1896 x 10 3 mol MnO  x 5 mol Fe2
4 4

1mol MnO 
4

 0.01095 mol

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4. convert mol calculated in step 3 to the
required units.

 Fe2  FeSO 4  0.01095
0.02L

 0.5474 M

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Exercise 2

How many mililitres of 0.112 M HCl will
react exactly with 21.2 mL ,0.150 M
sodium carbonate (Na2CO3)?
Answer : 56.8 mL

158

Solution

1. check/ write a balance equation.
2HCl + Na2CO3  2NaCl + H2CO3

0.112M 21.2 mL
? (M) 0.150 M

2. calculate the number of mol

mol Na 2CO3  0.150 x 21.2
1000

 3.18 x 10 3 mol

159

3. use the equation (mol ratio) to determine the
mole of unknown.

1 mol Na 2CO3  2 mol HCl

3.18 x 10 3 mol Na 2CO3  3.18 x103 mol x2 mol
1mol

 6.36 x 103 mol

160

4. convert mol calculated in step 3 to the
required units.

Volume of HCl  6.36 x103 mol x1000
0.112

 56.79 mL

161

Learning Outcome

At the end of this topic, students should be
able to:

a) define the limiting reactant and
percentage yield

b) perfome stoichiometric calculations
using mole concept including limiting
reactant and percentage yield.

162

Limiting reactant

• In practice, the reactants used usually not
present in exact stoichiometric amounts.

– Consequently, some reactant will be
used up while others will be left over
at the end of the reaction.

163

Limiting reactant & excess reactant

• Limiting reactant
A reactant that is completely consumed
in a reaction and limits the amount of
product formed.

• Excess reactant
Reactant that remains after reaction,
presents at greater quantity compared
to the limiting reactant.

164

Example 1 frame

Consider a bicycle: requires
2 tyres

Given: +

2 frames Five tyres

How many bicycles will be formed ?
What limits the product ? The frame (limiting reactant)

165

Example

• Consider the reaction:

Zn + 2HCl ZnCl2 + H2

• From equation:
1 mol Zn reacts completely with 2 mol HCl

• If 2 mole of Zn were added to 3 mole of HCl
- Limiting reactant : HCl
- Reactant in excess: Zn

166

Steps to determine limiting reactant

i) Calculate no of moles available for each reactant
ii) Stoichiometrically, calculate no of moles of

reactant
iii) Compare no of mol calculated (needed) in (ii) to

no of moles available in (i)
mol needed > mol available limiting reactant
mol needed < mol available excess reactant
iv) Apply no of moles of limiting reactant to
determine the product.

167

Example1

A 10.0 g of hydrochloric acid was reacted
with 10.0 g of calcium carbonate to
produce carbon dioxide gas. Determine
a) the limiting reactant
b) the mass of CO2 gas produced
c) the mass of unused reactant

168

Solution:
2HCl + CaCO3  CaCl2 + H2O + CO2

10.0 g 10.0 g

a)
10.0 g

mol of HCl = 36.46 g mol-1
= 0.274 mol

10.0 g
mol of CaCO3 = 100.09 g mol-1

= 0.0999 mol

169

from equation: 2 mol HCl  1 mol CaCO3

The number of moles of CaCO3 needed to react with
all HCl is

mol of CaCO3= 0.274 mol HCl x 1 mol CaCO3
2 mol HCl

= 0.137 mol CaCO3

Only 0.0999 mol CaCO3 available. Thus, not
enough CaCO3 to react with all HCl.

∴ CaCO3 is the limiting reagent.

170

b) The mass of CO2 produced depends on
the amount of the limiting reactant.

from equation: 1 mol CaCO3  1 mol CO2

mole of CO2 = mole of CaCO3
= 0.0999 mol

mass of CO2 = 0.0999 mol x 44.01 g mol-1
= 4.40 g

171

c) HCl is the excess reactant

from equation: 1 mol CaCO3  2 mol HCl

mole of HCl used = 0.0999 mol CaCO3x 1 2 mol HCl
mol CaCO3

= 0.200 mol

mass of HCl used = 0.200 mol x 36.46 g mol-1
= 7.29 g

172

Mass of HCl available = 10.0 g
∴ Mass of HCl unused = 10.0 – 7.29 g

= 2.71 g

173

Example 2

A 3.200 g sample of NaOH is added to a
solution containing 1.125 g of H2SO4.
Determine
a) the limiting reactant
b) the mass of sodium sulphate formed
c) the mass of unused reactant

174

Solution:

2NaOH + H2SO4  Na2SO4 + 2H2O

3.200 g 1.125 g

a)

3.200 g
mol of NaOH = 40.00 g mol-1

= 0.08000 mol

1.125 g
mol of H2SO4 = 98.08 g mol-1

= 0.01147 mol

175

from equation: 2 mol NaOH  1 mol H2SO4
Moles of H2SO4 needed to react with all NaOH,

mol of H2SO4 = 0.08000 mol NaOH x 1 mol H2SO4
2 mol NaOH

= 0.04000 mol

Only 0.01147 mol H2SO4 available. Thus, not
enough H2SO4 to react with all NaOH.

∴ H2SO4 is the limiting reactant.

176

b)

from equation: 1 mol H2SO4  1 mol Na2SO4

mole of Na2SO4 = mole of H2SO4
= 0.01147 mol

mass of Na2SO4 = 0.01147 mol x 142.05 g mol-1
= 1.629 g

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c) The excess reactant = NaOH

from equation: 1 mol H2SO4  2 mol NaOH

2 mol NaOH
mole of NaOH used = 0.01147 mol H2SO4x 1 mol H2SO4

= 0.02294 mol

mass of NaOH used = 0.02294 mol x 40.00 g mol-1
= 0.9176 g

178

Mass of NaOH available = 3.200 g
∴ Mass of NaOH unused = 3.200 – 0.9176 g

= 2.282 g

179

Percent yield

• Theoretical yield
The amount of product predicted to form
based on the balanced equation.
(based on stoichiometric calculation)

• Actual yield
The amount of product actually obtained
from a reaction in an experiment.

180

• Percent yield

% yield = actual yield x 100
theoretical yield

181

Example

Titanium is prepared by the reaction of
titanium(IV) choride with molten magnesium
between 950oC and 1150oC.

TiCl4(g) + 2Mg(l) → Ti(s) + 2MgCl2(l)

In a certain industrial operation 3.54 X 107 g
of TiCl4 are reacted with 1.13 x 107g of Mg.

a) Calculate the theoretical yield of Ti in
grams

b) Calculate the percent yield if 7.91 X 106 g

of Ti are actually obtained.

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Solution

a)

3.54 x 107g
nTiCl4 = 189.7 g mol-1

= 1.87 x 105 mol

n Mg = 1.13 x 107 g
24.31 g mol-1

= 4.65x105 mol

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From equation: 1 mol TiCl4  2 mol Mg

mol of Mg = 1.87 x 105mol TiCl4 x 2 mol Mg
1 mol TiCl4

= 3.74 x 105 mol

4.65 x 105 mol of Mg available. Thus, Mg is the
excess reactant.

∴ TiCl4 is the limiting reactant.

184

From equation: 1 mol TiCl4  1 mol Ti

mole of Ti = mole of TiCl4
= 1.87x 105 mol

The theoretical = 1.87 x 105 mol x 47.88g/mol
mass of Ti formed = 8.95 X 106 g

185

b)

Actual yield x 100%

% Yield = Theoretical yield

= 7.91 x 106 g x 100%
8.93 x 106 g

= 88.6 %

186

Exercise

Consider the reaction:
2Sb (s) + 3I2 (s)  2SbI3 (s)

Determine the limiting reactant and the theoretical
yield when 1.20g of Sb and 2.40g I2 are mixed.
What mass of excess reactant is left when the
reaction is complete.
( Sb=121.8, I= 126.9)

187

Solution:

n Sb  1.20 n I2  2.40
121.8 126.9x2

 9.8522 x103 mol  9.4563 x103 mol

(given ) (given )

2mol Sb  3 mol I2
9.8522 x 103 mol Sb  9.8522 x 103 x 3

2
 0.01479 mol I2 (needed )

Mole needed I2 > mole given I2
Thus, I2 is limiting reactant

188

Solution:

3 mol I2  2 mol SbI3

9.4563 x 10 3 mol I2  9.4563 x 10 3 x2
3

theoretica l yield  6.3042 x103 mol SbI3

189

3mol I2  2 molSb

9.4563 x 103 molI2  9.4563 x 103 x2
3

moleneeded Sb  6.3042 x103 molSb

moleSb unreacted  9.8522 x103  6.3042 x103
 3.548 x103 mol

massSb unreacted  3.548 x103 mol x121.8 gmol1
 0.4321g

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