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?

Question

Frederick Graves

Chemistry

3 June, 18:37

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?

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Skyler Franklin · 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