Lesson 9 Mole Ratios and Theoretical Yields

Refresh 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)? 12.5 25.0 50.0 75.0

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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 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

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!

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

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? 2 C4H10 + 13 O2  8 CO2 + 10 H2O C4H10 is given, O2 is wanted. n(O2) = 0.215 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

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

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 = 0.125 (smallest therefore is L.R.) Given is HCl, wanted is MgCl2 n(MgCl2) = (0.100 x 2.50) x (1 / 2) n(MgCl2) = 0.125 mol Determine limiting reagent Assign ‘wanted’ and ‘given’ State the equation Sub in your numbers Evaluate the sum

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 = 12.00 – (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

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.

AgNO3(aq) + NaCl(aq)  AgCl(aq) + NaNO3(aq) 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) Determine theoretical yield: AgCl is wanted, AgNO3 is given n(AgCl) = 0.150 x (1/1) = 0.150 mol % Yield = 0.125 / 0.150 x 100 = 83.3% Check for balanced equation Calculate theoretical yield using previous maths Determine % yield

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 0.437 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 = 0.565 mol Check for balanced equation Rearrange yield equation 𝑡ℎ𝑒𝑜𝑟𝑒𝑡𝑖𝑐𝑎𝑙 𝑦𝑖𝑒𝑙𝑑= 𝑎𝑐𝑡𝑢𝑎𝑙 𝑦𝑖𝑒𝑙𝑑 % 𝑦𝑖𝑒𝑙𝑑 ×100 Sub-in the numbers

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