Download Mole Ratios and Theoretical Yields

Lesson 9
Mole Ratios and Theoretical Yields
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
What volume, in cm3, of 0.200 mol dm–3 HCl(aq) is
required to neutralize 25.0 cm3 of 0.200 mol dm–3
Ba(OH)2(aq)?
A.
12.5
B.
25.0
C.
50.0
D.
75.0
We Are Here
Lesson 9: Mole Ratios and Theoretical Yields

Objectives:

Know how to construct and balance symbol equations

Apply the concept of the mole ratio to determine the amounts
of species involved in chemical reactions

Meet the idea of the ‘limiting reagent’
Mole Ratios

This is the ratio of one compound to another in a balanced
equation.

For example, in the previous equation
2 H2 + O2  2 H2O
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Hydrogen, oxygen and water are present in 2:1:2 ratio.
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This ratio is fixed and means for example:
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

0.2 mol of H2 reacts with 0.1 mol of O2 to make 0.2 mol H2O
5 mol of H2 reacts with 2.5 mol of O2 to make 5 mol of H2O
To make 4 mol of H2O you need 4 mol of H2 and 2 mol of O2
Mole Ratios in Calculations

You will often have questions that ask you how many moles of X can
be made from a amount of Y
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Or various similar questions
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Use the following:
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Where:


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The mole ratio!
wanted = the substance you want to find out more about
given = the substance you are given the full info for
n(wanted) = the number of moles you are trying to find out
n(given) = the number of moles of you are given in the question
wanteds = the number of wants in the balanced equation
givens = the number of givens in the balanced equation
Example 1…

What quantity of Al(OH)3 in moles is required
to produce 5.00 mol of H2O?
2 Al(OH)3 + 3 H2SO4  Al2(SO4)3 + 3 H2O

H2O is given, Al(OH)3 is wanted.
Check for balanced
equation
Assign ‘wanted’
and ‘given’
State the
equation


n(Al(OH)3) = 5.00 x (2/3)
Sub in your
numbers

n(Al(OH)3) = 3.33 mol
Evaluate the
sum
Example 2…you try

What quantity of O2 in moles is required to
fully react with 0.215 mol of butane (C4H10) to
produce water and carbon dioxide?
Check for balanced
equation
2 C4H10 + 13 O2  8 CO2 + 10 H2O

C4H10 is given, O2 is wanted.
Assign ‘wanted’
and ‘given’
State the
equation
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
n(O2) = 0.215 x (13/2)

n(O2) = 1.40 mol
Sub in your
numbers
Evaluate the
sum
The Limiting Reagent

In a reaction, we can describe reactants as being ‘limiting’ or in ‘excess’


Limiting – this is the reactant that runs out
Excess – the reaction will not run out of this
2 H2 + O2  2 H2O
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For example, if you have 2.0 mol H2 and 2.0 mol O2
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
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To determine this, divide the quantity of each reactant by its coefficient in the
equation. The smallest number is the limiting reactant:
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H2 is the limiting reactant – it will run out
O2 is present in excess – there is more than enough
H2: 2.0 / 2 = 1.0 – smallest therefore limiting
O2: 2.0 / 1 = 2.0
The limiting reactant will be your ‘given’ in all further calculations:

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Determining amounts of products formed
Determining amounts of other reactants used
Example 1: What quantity, in moles, of MgCl2 can be produced by reacting
10.5 g magnesium with 100 cm3 of 2.50 mol dm-3 hydrochloric acid solution?

Mg + 2HCl  MgCl2 + H
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Determining limiting reagent:
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Mg: (10.5 / 24.31)/1 = 0.432
HCl: (0.100 x 2.50)/2 = 0.125 (smallest therefore is L.R.)
Given is HCl, wanted is MgCl2
Check for balanced
equation
Determine limiting
reagent
Assign ‘wanted’
and ‘given’
State the
equation
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n(MgCl2) = (0.100 x 2.50) x (1 / 2)
Sub in your
numbers
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n(MgCl2) = 0.125 mol
Evaluate the
sum
Example 2 (you try): What quantity, in moles, of carbon dioxide would be
formed from the reaction of 12.0 mol oxygen with 2.00 mol propane, and
how much of which reactant would remain?
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C3H8 + 5O2  3CO2 + 4H2O
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Determining limiting reagent:
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C3H8: 2.00 / 1 = 2.00 (smallest therefore is L.R.)
O2: 12.0 / 5 = 2.40
Given is C3H8, wanted is CO2
Check for balanced
equation
Determine limiting
reagent
Assign ‘wanted’
and ‘given’
State the
equation
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n(CO2) = 2.00 x (3 / 1) = 6.00 mol
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N(O2) remaining
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= n(O2) at start – n(O2) used
= 12.00 – (2.00 x (5 / 1) )
= 2.00 mol
Sub in your
numbers
Evaluate the
sum
Theoretical, actual and percentage yield
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Theoretical yield is the maximum amount of product you would make if
the limiting reactant was fully converted to product.
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Use the limiting reactants maths to work this out
Actual yield is the actual amount of product collected in after a reaction
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It is always less than the theoretical yield
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Percentage yield reflects how close you got to achieving the theoretical
yield:
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Your actual and theoretical yields can be in either moles or grams, so long
as they are both the same units.
Example: 0.150 mol of silver nitrate was reacted with excess sodium chloride.
After filtration, 0.125 mol of silver chloride was collected. What was the %
yield?

AgNO3(aq) + NaCl(aq)  AgCl(aq) +
NaNO3(aq)
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Determine theoretical yield:
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AgCl is wanted, AgNO3 is given

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Calculate
theoretical yield
using previous
maths
n(AgCl) = 0.150 x (1/1) = 0.150 mol
Determine %
yield


Check for balanced
equation
% Yield = 0.125 / 0.150 x 100 = 83.3%
Example: After the thermal decomposition of some calcium carbonate, I
collected 0.437 mol of calcium oxide, which was a 77.4% yield. How much
calcium carbonate did I start with?

CaCO3  CaO + CO2
Check for balanced
equation

𝑎𝑐𝑡𝑢𝑎𝑙 𝑦𝑖𝑒𝑙𝑑
𝑡ℎ𝑒𝑜𝑟𝑒𝑡𝑖𝑐𝑎𝑙 𝑦𝑖𝑒𝑙𝑑 =
× 100
% 𝑦𝑖𝑒𝑙𝑑


theoretical yield
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= (0.437 / 77.4) x 100
= 0.565 mol
Rearrange yield
equation
Sub-in the
numbers
Key Points
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Balance equations by playing with coefficients
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Use mole ratios to work quantities of chemicals involved
in reactions
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Divide the quantity of each reactant by its coefficient to
determine the limiting reactant