1. Lesson 9
Mole Ratios and Theoretical Yields

2. Refresh
What volume, in cm3, of mol dm–3 HCl(aq) is required to neutralize 25.0 cm3 of mol dm–3 Ba(OH)2(aq)?
12.5
25.0
50.0
75.0

3. We Are Here

4. 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’

5. 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
Hydrogen, oxygen and water are present in 2:1:2 ratio.
This ratio is fixed and means for example:
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

6. 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
Or various similar questions
Use the following:
Where:
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
The mole ratio!

7. 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.
n(Al(OH)3) = 5.00 x (2/3)
n(Al(OH)3) = 3.33 mol
Check for balanced equation
Assign ‘wanted’ and ‘given’
State the equation
Sub in your numbers
Evaluate the sum

8. Example 2…you try
What quantity of O2 in moles is required to fully react with mol of butane (C4H10) to produce water and carbon dioxide?
2 C4H O2  8 CO H2O
C4H10 is given, O2 is wanted.
n(O2) = x (13/2)
n(O2) = 1.40 mol
Check for balanced equation
Assign ‘wanted’ and ‘given’
State the equation
Sub in your numbers
Evaluate the sum

9. 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
For example, if you have 2.0 mol H2 and 2.0 mol O2
H2 is the limiting reactant – it will run out
O2 is present in excess – there is more than enough
To determine this, divide the quantity of each reactant by its coefficient in the equation. The smallest number is the limiting reactant:
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:
Determining amounts of products formed
Determining amounts of other reactants used

10. Determining limiting reagent:
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?
Check for balanced equation
Mg + 2HCl  MgCl2 + H
Determining limiting reagent:
Mg: (10.5 / 24.31)/1 = 0.432
HCl: (0.100 x 2.50)/2 = (smallest therefore is L.R.)
Given is HCl, wanted is MgCl2
n(MgCl2) = (0.100 x 2.50) x (1 / 2)
n(MgCl2) = mol
Determine limiting reagent
Assign ‘wanted’ and ‘given’
State the equation
Sub in your numbers
Evaluate the sum

11. Determining limiting reagent:
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?
C3H8 + 5O2  3CO2 + 4H2O
Determining limiting reagent:
C3H8: 2.00 / 1 = 2.00 (smallest therefore is L.R.)
O2: 12.0 / 5 = 2.40
Given is C3H8, wanted is CO2
n(CO2) = 2.00 x (3 / 1) = 6.00 mol
N(O2) remaining
= n(O2) at start – n(O2) used
= – (2.00 x (5 / 1) )
= 2.00 mol
Check for balanced equation
Determine limiting reagent
Assign ‘wanted’ and ‘given’
State the equation
Sub in your numbers
Evaluate the sum

12. Theoretical, actual and percentage yield
Theoretical yield is the maximum amount of product you would make if the limiting reactant was fully converted to product.
Use the limiting reactants maths to work this out
Actual yield is the actual amount of product collected in after a reaction
It is always less than the theoretical yield
Percentage yield reflects how close you got to achieving the theoretical yield:
Your actual and theoretical yields can be in either moles or grams, so long as they are both the same units.

13. AgNO3(aq) + NaCl(aq)  AgCl(aq) + NaNO3(aq)
Example: mol of silver nitrate was reacted with excess sodium chloride. After filtration, mol of silver chloride was collected. What was the % yield?
AgNO3(aq) + NaCl(aq)  AgCl(aq) + NaNO3(aq)
Determine theoretical yield:
AgCl is wanted, AgNO3 is given
n(AgCl) = x (1/1) = mol
% Yield = / x 100 = 83.3%
Check for balanced equation
Calculate theoretical yield using previous maths
Determine % yield

14. CaCO3  CaO + CO2 theoretical yield = (0.437 / 77.4) x 100 = 0.565 mol
Example: After the thermal decomposition of some calcium carbonate, I collected mol of calcium oxide, which was a 77.4% yield. How much calcium carbonate did I start with?
CaCO3  CaO + CO2
theoretical yield
= (0.437 / 77.4) x 100
= mol
Check for balanced equation
Rearrange yield equation
𝑡ℎ𝑒𝑜𝑟𝑒𝑡𝑖𝑐𝑎𝑙 𝑦𝑖𝑒𝑙𝑑= 𝑎𝑐𝑡𝑢𝑎𝑙 𝑦𝑖𝑒𝑙𝑑 % 𝑦𝑖𝑒𝑙𝑑 ×100
Sub-in the numbers

15. Key Points Balance equations by playing with coefficients
Use mole ratios to work quantities of chemicals involved in reactions
Divide the quantity of each reactant by its coefficient to determine the limiting reactant