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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.

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joseaaronlara

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

ajeigbeibraheem

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:

  • mass = 41.0 g
  • Molar mass = 24.31 g.mol

For FeCl₃:

  • mass = 175 g
  • Molar mass = 162.2 g/mol

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

[tex]\mathbf{Number \ of \ moles = \dfrac{mass}{molar \ mass}}[/tex]

For Mg:

  • [tex]\mathbf{Number \ of \ moles = \dfrac{41.0 \ g}{24.31 \ g/mol}}[/tex]
  • [tex]\mathbf{Number \ of \ moles =1.69 \ moles}[/tex]

For FeCl₃:

  • [tex]\mathbf{Number of moles = \dfrac{175 \ g}{162.2 \ g/mol}}[/tex]
  • [tex]\mathbf{Number \ of \ moles =1.08 \ moles}[/tex]

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;

  • 3(24.31) grams of Mg reacts with 2(162.2) grams of FeCl₃

Similarly;

  • (x) gram of Mg will react with 175 g of FeCl₃

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

  • [tex]\mathbf{x = \dfrac{3(24.31) \times 175 g }{2(162.2) g }}[/tex]
  • x = 39.34 grams

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.

Learn more about the chemical reaction here:

https://brainly.com/question/1689737?referrer=searchResults

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