Transcript File - Ciencias esmeralda

Chemical quantities in reactions
What and why
 If we know the chemical equation for a product:
 We can work out how much we need
 We can work out how much we will get
 We can work out the percentage efficiency of
processes
 Why is all this so important in the real world?
Starter questions
 1. What is the mass of 0.5moles of water?
 2. What are the products of this equation?
 Ca + 2HCl =
 3. Balance this equation
 Na + Cu3(PO4)2 = Na3PO4 + Cu
 4. Complete and balance this double substitution
reaction.
 Ag2SO4 + Pb(NO3)4 =
Calculations in equations
 Calculate the mass of reactants and products when
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1mole of C burns in oxygen.
(What do we know about the mass of reactants and
products from the start?)
C + O2 = CO2
Is this balanced?
Molar masses C = 12g, O2 = 32g
12 + 32 = 44g = 1 mole of CO2
Amount of reagents is 44g. Amount of products is
44g
Do these
 2 moles Iron + 3 moles Sulphur
 2 mole Sodium hydroxide + 2 mole Hydrochloric
acid
Moles of a reactant needed.
 How many moles of sulphur are needed to react with
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1.42 moles of iron (iii)?
Write equation
2Fe + 3S = Fe2S3
Ratio of Iron to Sulphur is 2:3
3/2 x amount of iron = amount of sulphur
1.5 x 1.42 = 2.13moles = amount of sulphur
2.13 x 32g = 68.16g = mass of sulphur needed.
Limiting reactants
 If you don’t have enough of one reactant it will limit how
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much product is produced.
Making hydrogen chloride
Equation H2 + Cl2 = 2HCl
So if we have one mole of each reactant we can get 2
moles of the product
But if we have 2 moles of hydrogen and 5 moles of
chlorine how much can we get?
Which is limiting
How much product can we make per mole?
So how much product can we make in total?
Limiting masses
 Ammonia and fluorine react to produce dinitrogen
tetrafluoride and hydrogen fluoride.
 NH3 + F2 = N2F4 + HF (balance this!)
 If 5g of NH3 and 20g of F2 react how many g of HF
are produced
 Steps:
Steps
 5 g NH3 Molar Mass, Moles NH3 mole-mole factor,
moles HF, Molar mass, grams HF
 20g F2 Molar mass, Moles F2 mole-mole factor,
moles of moles HF, molar mass, grams HF
 Molar mass of NH3 = 17g. We have 5g so we have
0.294 moles. 2 moles of NH3 make 6 moles of HF so
0.294 moles will make 3 x as much HF i.e. 0.882
moles
next
 Molar mass of F2 = 38g
 20g of F2 = 20/38 = 0.526moles of F2
 5 moles of F2 make 6 moles of HF
 So 0.526 moles of F2 will make 6/5 x 0.526 moles of
HF, i.e. 0.632 moles of HF
 This is fewer moles of HF than can be made from the
ammonia so the fluorine is the limiting factor.
 (I will post exercises on blackboard)
And finally
Percentage yields
 In reality no chemical reactions work perfectly.
 For this reason it is important to work out the
percentage yield by dividing the actual yield by the
theoretical yield you would get if it were 100%
efficient.
 Percentage yield = actual yield / theoretical
yield x 100.
Example
 Equation LiOH + CO2 = LiHCO3
 What is the % yield if 50g LIOH gives 72.8g LiHCO3
 Equation is balanced. 1 mole LiOH gives 1 mole
LiHCO3
 50g / molar mass of LiOH = 50 / 24 = 2.08moles.
As mole factor is 1:1 reactant to product then it
should give 2.08 moles of LiHCO3
Moving on
 1 mole of LiHCO3 has a mass of 68g
 2.08 moles of LiHCO3 has a mass of 141.67g
 % yield = actual yield /theoretical yield
 = 72.8/141.67 = 51.4%
There are plenty of activities to keep us busy!