File - Ciencias esmeralda
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
Why is all this so important in the real world?
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
Ag2SO4 + Pb(NO3)4 =
Calculations in equations
Calculate the mass of reactants and products when
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
2 moles Iron + 3 moles Sulphur
2 mole Sodium hydroxide + 2 mole Hydrochloric
Moles of a reactant needed.
How many moles of sulphur are needed to react with
1.42 moles of iron (iii)?
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.
If you don’t have enough of one reactant it will limit how
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?
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
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
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)
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%
Percentage yield = actual yield / theoretical
yield x 100.
Equation LiOH + CO2 = LiHCO3
What is the % yield if 50g LIOH gives 72.8g LiHCO3
Equation is balanced. 1 mole LiOH gives 1 mole
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
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!