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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(g) → 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 acid, the diagram below explains this:

reaction of magnesium and hydrochloric acid

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. 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 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 obtianed, 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 by burning 10g, so simply multiply 1.67 x 10= 16.7g of magnesium oxide produced.

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