Higher tier
Limiting reactions
Magnesium ribbon reacts with hydrochloric acid
according to the equations below:
magnesium(s) + hydrochloric acid(aq) → magnesium chloride(aq) + hydrogen (g)
Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g)
Most of the time when carrying out a chemical reaction it is common to use an excess of one
reactant (this simply means to use more than you actually need to react), this will ensure
that the other reactant is completely used up. In the reaction of magnesium
with hydrochloric acid,
to ensure that all the magnesium ribbon reacts use an excess
of the hydrochloric acid, the diagram below explains this:
The reactant which is
not in excess will be the one that will determine the maximum amount of
product that can be obtained- it is called the limiting factor or
limiting reactant e.g.
Magnesium ribbon burns in air to form magnesium oxide, here the air is
clearly in excess and
the limiting factor is the amount of magnesium ribbon. It is the mass of
magnesium that will determine
the amount of magnesium oxide produced.
To calculate the mass of magnesium oxide formed when a given mass of magnesium is burned
simply use the method shown here. In the exam just make sure
you
identify which of the reactants is in excess
as it is this one that will determine the maximum mass
of product or yield of reaction.
e.g. Calculate the maximum mass of magnesium oxide
obtained by burning 10g of magnesium ribbon in air.
step 1- You need to write a balanced symbolic equation for the reaction, more than likely you will be given
one in the exam!
magnesium(s) + oxygen(g) → magnesium oxide(s)
2Mg(s) + O2(g) → 2MgO(s)
Step2- Calculate the relative formula masses, we don't need to bother with oxygen as it is in excess. It is the amount
of magnesium that will determine the mass of the product obtained, it is the limiting factor here.
Ar of Mg=24 Ar of O=16
2Mg(s) → 2MgO(s)
Ar=2 x24=48 → Mr= 24 + 16=40
remember the Ar or Mr expressed in grams will give the mass of 1 mole of the substance. So here
2 moles or 48g of magnesium will produce 2 moles or 80g of
magnesium oxide.
However in this example we are only burning 10g of magnesium, not 48g. So simply scale down. I
would suggest to begin with you calculate the mass of magnesium oxide produced by burning 1g of
magnesium:
2Mg(s) → 2MgO(s)
48g → 80g
So divide both sides of the equation by 48, since any number divided by itself is 1,
this will give the mass of magnesium oxide produced by burning 1g of
magnesium.
2Mg(s) → 2MgO(s)
48g/48 → 80g/48
1g → 1.67g
So burning 1g of magnesium produces 1.67g of magnesium oxide, in this example the question
asked how much magnesium oxide is produced by burning 10g of
magnesium, so simply multiply 1.67 x 10= 16.7g of magnesium
oxide produced.
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