Question

When aluminum (Al) reacts with chlorine gas (Cl2), aluminum chloride (AlCl3) is produced according to the balanced equation below: 2Al + 3Cl2 → 2AlCl3 If 20 atoms of aluminum react with 45 molecules of chlorine gas, which reactant is limiting and how many more atoms/molecules would be required to use up all the reactants?

Answer 1

The balanced reaction shows for every 2 Aluminum we need 3 Chloride gas.

20 Al * (3 Cl / 2 Al) = 30 Cl
We have 45 Cl ... Which is more than enough for 20 Al
We have excess Cl
Then the limiting reagent =  Al

Answer 2

Answer : Aluminium atom is limiting reagent and 10 more atoms will be required.

Solution :

The balanced reaction is,

$2Al+3Cl_2\rightarrow 2AlCl_3$

According to the reaction, 2 atoms of aluminium react with 3 molecules of chlorine gas which means that 6 atoms of chlorine.

(1) Then 20 atoms of aluminum react with :

$\frac{6}{2}\times 20=60$ atoms of chlorine

60 atoms of chlorine = $\frac{60}{2}=30$ molecules of chlorine gas

(2) Now 45 molecules of chlorine gas will react with :

Number of atoms of chlorine in 45 molecules of chlorine gas = 90 atoms

Now 90 atoms of chlorine will react with : $\frac{2}{6}\times 90=30$ atoms of aluminium

From this we conclude that the aluminum is a limiting reagent and the chlorine is an excess reagent.

Number of aluminium atoms required = Number of aluminium required to consume chlorine - number of aluminium atoms given

Number of aluminium atoms required = 30 - 20 = 10 atoms of aluminium atoms.