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.

Magnesium (used in the manufacture of light alloys) reacts with iron(III) chloride to form magnesium chloride and iron.

The limiting reactant is FeCl₃ and the mass of the excess reactant present in the vessel when the reaction is complete is 1.66 g.

The chemical reaction equation is:

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

For Mg:

For FeCl₃:

Using the relation for calculating the number of moles, which is:

For Mg:

For FeCl₃:

To determine the limiting reactant, we will divide each reactant by its Stochiometric coefficient.

i.e.

For Mg:

= 1.69 moles/3

= 0.56

For FeCl₃:

= 1.08 moles / 2

= 0.54

Hence, FeCl₃ is the limiting reactant.

However, to determine the mass of the excess reactant present in the vessel when the reaction is complete, we have the following;

From the reaction, if 3 moles of Mg reacts with 2 moles FeCl₃

Then;

Similarly;

To determine the amount of (x) gram, we have:

Thus, the excess mass of Mg = 41.0 g - 39.34 g

= 1.66 grams

In conclusion, the limiting reactant is FeCl₃ and the mass of the excess reactant present in the vessel when the reaction is complete is 1.66 g.

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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

1 a

2 d

3 a

4 c

explanation: