Answer to Question #126133 in General Chemistry for Amanda

Question

For the reaction shown, find the limiting reactant for each of the following initial amount of reactants.
2Al (s)+3Cl2 (g) —> 2AlCl3 (s)
1.0g Al; 1.0g Cl2

Answer 1

M(Al) = 26.98 g/mol;

n=m/M;

n(Al) = 1/26.98 = 0.037 moles;

M(Cl₂) = 70.90 g/mol;

n(Cl₂) = 1/70.90 = 0.014 moles;

The mole ratio Al:Cl₂ according to the reaction is 2:3;

The mole ratio Al:Cl₂ according to the initial amount of reactants is 0.037:0.014 or 2.6:1.

Answer: Cl₂ is the limiting reactant.