Question

Translate the following statement into chemical equations and balance:

Barium chloride reacts with aluminium sulphate to give aluminium chloride and precipitate of barium sulphate.

Answer 1

Barium chloride reacts with aluminium sulphate to give aluminium chloride and precipitate of barium sulphate.

1.Reaction:

$\mathrm{BaCl}{}_2+\mathrm{Al}{}_2(\mathrm{SO}{}_4){}_3\to \mathrm{BaSO}{}_4+\mathrm{AlCl}{}_3$

(Barium chloride) (Aluminium sulphate) (Barium sulphate) (Aluminium chloride)

2. Balancing steps

Step 1: Write unbalanced equation.

$\mathrm{BaCl}{}_2+\mathrm{Al}{}_2(\mathrm{SO}{}_4){}_3\to \mathrm{BaSO}{}_4+\mathrm{AlCl}{}_3$

Step 2: Write down the number of atoms of each element present on both reactant as well as product side.

Element Reactant side Product side

$\mathrm{Ba}$ $1$ $1$

$\mathrm{Cl}$ $2$ $3$

$\mathrm{Al}$ $2$ $1$

$\mathrm{S}$ $3$ $1$

$\mathrm{O}$ $12$ $4$

Step 3: Add coefficients to balance the number of atoms at both the sides

Firstly write $2$ before $\mathrm{AlCl}{}_3$

$\mathrm{BaCl}{}_2+\mathrm{Al}{}_2(\mathrm{SO}{}_4){}_3\to \mathrm{BaSO}{}_4+2\mathrm{AlCl}{}_3$

Secondly add $3$ before $\mathrm{BaCl}{}_2$

$3\mathrm{BaCl}{}_2+\mathrm{Al}{}_2(\mathrm{SO}{}_4){}_3\to \mathrm{BaSO}{}_4+2\mathrm{AlCl}{}_3$

Thirdly add $3$ before $\mathrm{BaSO}{}_4$

$3\mathrm{BaCl}{}_2+\mathrm{Al}{}_2(\mathrm{SO}{}_4){}_3\to 3\mathrm{BaSO}{}_4+2\mathrm{AlCl}{}_3$

Balanced equation is:

$3\mathrm{BaCl}{}_2(\mathrm{aq})+\mathrm{Al}{}_2(\mathrm{SO}{}_4){}_3(\mathrm{aq})\to 3\mathrm{BaSO}{}_4(\mathrm{aq})+2\mathrm{AlCl}{}_3(\mathrm{aq})$

(Barium chloride) (Aluminium sulphate) (Barium sulphate) (Aluminium chloride)