Limiting reactants - (higher tier)
A reaction finishes when one of the reactantA substance that reacts together with another substance to form products during a chemical reaction. is all used up. The other reactant has nothing left to react with, so some of it is left over:
- the reactant that is all used up is called the limiting reactantThe reacting substance that is completely used up in a chemical reaction and which determines how much product is made. - it sets a limit on how much productsSubstances formed in a chemical reaction. can form.
- the reactant that is left over is described as being in excessIn chemistry, a substance is in excess if there is more than enough of it to react with another reactant.
The massThe amount of matter an object contains. Mass is measured in kilograms (kg) or grams (g). of product formed in a reaction depends upon the mass of the limiting reactant. This is because no more product can form when the limiting reactant is all used up.
Example
How much aluminium chloride is formed when 2.7 g of aluminium are reacted with 9.125 g of HCl?
What mass of aluminium chloride is formed when 2.7g of aluminium are reacted with 9.125g of HCl?
2Al + 6HCl 鈫 2AlCl3 + 3H2
Convert massThe amount of matter an object contains. Mass is measured in grams (g). into mole The amount of substance that contains the same number of particles as there are atoms in 12 g of carbon-12. for both reactants:
\({moles~of~Al} = \frac{mass~(g)}{M_r} = \frac{2.7}{27} = 0.1\)
\({moles~of~HCl} = \frac{mass~(g)}{M_r} = \frac{9.125}{36.5} = 0.25\)
Now look at the ratio of the reactants:
2 mol Al : 6 mol HCl (1 mol Al : 3 mol HCl)
0.1 mol Al : 0.3 mole HCl
There are only 0.25 moles of HCl (instead of 0.3 moles), so the HCl will run out first. It is the limiting reactant.
Now use the moles of the limiting reactant to calculate the mass of the product. Remember to use the molar ratio between the limiting reactant and the product.
Moles of HCl = 0.25
6 mol HCl : 2 mol AlCl3 which simplifies to
3 mol HCl : 1 mol AlCl3
therefore:
0.25 mol HCl will make 0.0833 mol AlCl3.
Mass of AlCl3 = moles x Mr = 0.0833 x 133.5 = 11.12 g