If you react 25.0g of Cu with 25.0g of AlCl3 in the following reaction 3Cy + 2AlCl3 -> 3CuCl2 + 2Al a. find the excess and limiting reactants
b. calculate the mass of leftover reactant

a. AlCl₃ ⇒ limiting reactant(smaller ratio)

Cu ⇒ excess reactant

b. the mass of leftover reactant : 7.207 g

Given

25 g Cu

25 g AlCl3

Required

a. the excess and limiting reactants

b.  the mass of leftover reactant

Solution

Reaction

3Cu + 2AlCl₃ ⇒ 3CuCl₂ + 2Al

mol Cu(Ar = 63.5 g/mol) :

mol = mass : Mw

mol = 25 : 63.5

mol = 0.394

mol AlCl3(MW=133,34 g/mol) :

mol = 25 : 133,34 g/mol

mol = 0.187

mol ratio to reaction coefficient Cu : AlCl₃ =

AlCl₃ ⇒ limiting reactant(smaller ratio)

Cu ⇒ excess reactant

b. the mass of leftover reactant :

mol Cu = 3/2 x 0.187 = 0.2805

mol left = 0.394 - 0.2805 = 0.1135

mass = 0.1135 x 63.5 = 7.207 g

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