Question

Magnesium (used in the manufacture of light alloys) reacts with iron(III) chloride to form magnesium chloride and iron. 3Mg(s) + 2FeCl₃(s) → 3MgCl₂(s) + 2Fe(s) A mixture of 41.0 g of magnesium ( = 24.31 g/mol) and 175 g of iron(III) chloride ( = 162.2 g/mol) is allowed to react. Identify the limiting reactant and determine the mass of the excess reactant present in the vessel when the reaction is complete.

Answer 1

Answer:

Theanswer to your question is:

Limiting reactant = FeCl₃

Excess reactant = 1.66 g of Mg

Explanation:

Data

Mg = 41 g   = 24.31 g/mol

FeCl₃ = 175 g = 162.2 g/mol

                         3Mg(s) + 2FeCl₃(s) → 3MgCl₂(s) + 2Fe(s)

                      3(24.31) of Mg ------------------  2(162.2)  of FeCl₃

                      72.93 g of Mg ------------------ 324.4 g of FeCl₃

Theoretical Proportion = 324.4/72.93 = 4.44

Practical proportion   =  175 / 41 = 4.2

As the proportion disminishes the limiting reactant is FeCl₃.

Excess reactant

                               72.93 g of Mg ------------------ 324.4 g of FeCl₃

                                   x -------------------------           175 g of FeCl₃

x = (175 x 72.93) / 324.4

x = 39.34 g of Mg

Excess = 41 - 39.34

           = 1.66 g of Mg