Magnesium (used in the manufacture of light alloys) reacts with iron(III) chloride to form magnesium chloride and iron.
3Mg(s) + 2FeCl₃(s) $\rightarrow$ 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.

Question

Magnesium (used in the manufacture of light alloys) reacts with iron(III) chloride to form magnesium chloride and iron.
3Mg(s) + 2FeCl₃(s)

\rightarrow

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.

Quizplus · Accepted Answer

The Answer of Magnesium (used in the manufacture of light alloys) reacts with iron(III) chloride to form magnesium chloride and iron. 3Mg(s) + 2FeCl₃(s) $\rightarrow$ 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)