Question

Write the complete balanced reaction (a) Potassium bromide(aq) + Barium iodide(aq) → Potassium iodide(aq) + Barium bromide(s)

Answer 1

Balanced chemical equation: A balanced chemical equation is the representation of a chemical reaction in which the number of atoms of each element is the same on both the reactant and product sides.

Word equation to symbolic equation:

• In the given reaction, Potassium bromide reacts with Barium iodide to form Potassium iodide and Barium bromide.
• The chemical reaction can be represented as:

$\underset{\mathrm{Potassium}\mathrm{bromide}}{\mathrm{KBr}\left(\mathrm{aq}\right)}+\underset{\mathrm{Barium}\mathrm{iodide}}{{\mathrm{BaI}}_2\left(\mathrm{aq}\right)}\to \underset{\mathrm{Potassium}\mathrm{iodide}}{\mathrm{KI}\left(\mathrm{aq}\right)}+\underset{\mathrm{Barium}\mathrm{bromide}}{{\mathrm{BaBr}}_2\left(\mathrm{s}\right)}$

Balancing the Chemical Equation:

Step 1: Write the number of atoms of each element

$\underset{\mathrm{Potassium}\mathrm{bromide}}{\mathrm{KBr}\left(\mathrm{aq}\right)}+\underset{\mathrm{Barium}\mathrm{iodide}}{{\mathrm{BaI}}_2\left(\mathrm{aq}\right)}\to \underset{\mathrm{Potassium}\mathrm{iodide}}{\mathrm{KI}\left(\mathrm{aq}\right)}+\underset{\mathrm{Barium}\mathrm{bromide}}{{\mathrm{BaBr}}_2\left(\mathrm{s}\right)}$

Element	reactant side	product side
K	1	1
Br	1	2
Ba	1	1
I	2	1

• From the table, it is clear that the atoms of bromine and iodine are not balanced.

Step 2: Balancing the unbalanced atoms

• For balancing the Iodine atoms we will make the coefficient of $\mathrm{Kl}$ as $2$
• Now the equation becomes

$\underset{\mathrm{Potassium}\mathrm{bromide}}{\mathrm{KBr}\left(\mathrm{aq}\right)}+\underset{\mathrm{Barium}\mathrm{iodide}}{{\mathrm{BaI}}_2\left(\mathrm{aq}\right)}\to \underset{\mathrm{Potassium}\mathrm{iodide}}{2\mathrm{KI}\left(\mathrm{aq}\right)}+\underset{\mathrm{Barium}\mathrm{bromide}}{{\mathrm{BaBr}}_2\left(\mathrm{s}\right)}$

Element	Reactant side	Product side
K	1	2
Br	1	2
Ba	1	1
I	2	2

Now for balancing the rest of the atoms we will make the coefficient of $\mathrm{KBr}$ as $2$.

The equation becomes,
$\underset{\mathrm{Potassium}\mathrm{bromide}}{2\mathrm{KBr}\left(\mathrm{aq}\right)}+\underset{\mathrm{Barium}\mathrm{iodide}}{{\mathrm{BaI}}_2\left(\mathrm{aq}\right)}\to \underset{\mathrm{Potassium}\mathrm{iodide}}{2\mathrm{KI}\left(\mathrm{aq}\right)}+\underset{\mathrm{Barium}\mathrm{bromide}}{{\mathrm{BaBr}}_2\left(\mathrm{s}\right)}$

Element	Reactant side	Product side
K	2	2
Br	2	2
Ba	1	1
I	2	2

Hence, the balanced chemical equation is $\underset{\mathrm{Potassium}\mathrm{bromide}}{2\mathrm{KBr}\left(\mathrm{aq}\right)}+\underset{\mathrm{Barium}\mathrm{iodide}}{{\mathrm{BaI}}_2\left(\mathrm{aq}\right)}\to \underset{\mathrm{Potassium}\mathrm{iodide}}{2\mathrm{KI}\left(\mathrm{aq}\right)}+\underset{\mathrm{Barium}\mathrm{bromide}}{{\mathrm{BaBr}}_2\left(\mathrm{s}\right)}$