Question

Balance the following chemical equations and identify the type of chemical reaction.

${\mathrm{TiCl}}_4\left(\mathrm{l}\right)+\mathrm{Mg}\left(\mathrm{s}\right)\to \mathrm{Ti}\left(\mathrm{s}\right)+{\mathrm{MgCl}}_2\left(\mathrm{s}\right)$

Answer 1

Step 1: Given equation

${\mathrm{TiCl}}_4\left(\mathrm{l}\right)+\mathrm{Mg}\left(\mathrm{s}\right)\to \mathrm{Ti}\left(\mathrm{s}\right)+{\mathrm{MgCl}}_2\left(\mathrm{s}\right)$

Step 2: Method used for balancing the equation

Using the traditional method of balancing chemical equations. According to the method:

• First, compare the total number of atoms of each element on the reactant side and the product side.
• Add stoichiometric coefficients to molecules containing an element that has a different number of atoms on the reactant side and the product side.
• The coefficient must balance the number of atoms on each side.
• Usually, the stoichiometric coefficients for hydrogen and oxygen atoms are assigned last.
• The number of atoms of an element in one species must be obtained by multiplying the stoichiometric coefficient by the total number of atoms of that element present in 1 molecule of the species.
• Repeat this process till the number of all the atoms of the reacting elements are equal on both the reactant and product sides.

Step 3: Balancing the given equation

The given equation is: ${\mathrm{TiCl}}_4\left(\mathrm{l}\right)+\mathrm{Mg}\left(\mathrm{s}\right)\to \mathrm{Ti}\left(\mathrm{s}\right)+{\mathrm{MgCl}}_2\left(\mathrm{s}\right)$

Equation: ${\mathrm{TiCl}}_4\left(\mathrm{l}\right)+\mathrm{Mg}\left(\mathrm{s}\right)\to \mathrm{Ti}\left(\mathrm{s}\right)+{\mathrm{MgCl}}_2\left(\mathrm{s}\right)$
Reactant Side	Product Side
1 Titanium atom	1 Titanium atom
4 Chlorine atoms	2 Chlorine atoms
1 Magnesium atom	1 Magnesium atom

The Titanium atoms are already balanced.

Then the Chlorine atoms are balanced. After balancing the equation becomes: ${\mathrm{TiCl}}_4\left(\mathrm{l}\right)+\mathrm{Mg}\left(\mathrm{s}\right)\to \mathrm{Ti}\left(\mathrm{s}\right)+2{\mathrm{MgCl}}_2\left(\mathrm{s}\right)$

Finally, the magnesium atoms are balanced. After balancing the equation becomes: ${\mathrm{TiCl}}_4\left(\mathrm{l}\right)+2\mathrm{Mg}\left(\mathrm{s}\right)\to \mathrm{Ti}\left(\mathrm{s}\right)+2{\mathrm{MgCl}}_2\left(\mathrm{s}\right)$

The updated equation is: ${\mathrm{TiCl}}_4\left(\mathrm{l}\right)+2\mathrm{Mg}\left(\mathrm{s}\right)\to \mathrm{Ti}\left(\mathrm{s}\right)+2{\mathrm{MgCl}}_2\left(\mathrm{s}\right)$

Equation: ${\mathrm{TiCl}}_4\left(\mathrm{l}\right)+2\mathrm{Mg}\left(\mathrm{s}\right)\to \mathrm{Ti}\left(\mathrm{s}\right)+2{\mathrm{MgCl}}_2\left(\mathrm{s}\right)$
Reactant Side	Product Side
1 Titanium atom	1 Titanium atom
4 Chlorine atoms	4 Chlorine atoms
2 Magnesium atoms	2 Magnesium atoms

Since the number of all the atoms of the reacting elements are equal on both the reactant and product sides are equal, the equation is balanced.

Displacement Reaction

A displacement reaction is a reaction in which the atom or a set of atoms is displaced by another atom in a molecule. In this reaction, the titanium atom is displaced by the magnesium atom. Hence this reaction is a displacement reaction.

Hence the balanced equation is: ${\mathrm{TiCl}}_4\left(\mathrm{l}\right)+2\mathrm{Mg}\left(\mathrm{s}\right)\to \mathrm{Ti}\left(\mathrm{s}\right)+2{\mathrm{MgCl}}_2\left(\mathrm{s}\right)$ and this reaction is a displacement reaction.