Question

Write balanced equation for the following reaction:

(c) Iron(III) chloride solution with sodium hydroxide solution.

Answer 1

Reaction summary:

• When Iron (III) chloride $\left({\mathrm{FeCl}}_3\right)$ and sodium hydroxide $\left(\mathrm{NaOH}\right)$ undergoes a double decomposition reaction to give one insoluble compound in the form of precipitate ferric hydroxide $\left(\mathrm{Fe}{\left(\mathrm{OH}\right)}_3\right)$ and another soluble compound Sodium chloride$\left(\mathrm{NaCl}\right)$.

Skeleton reaction:

${\mathrm{FeCl}}_3\left(\mathrm{aq}\right)+\mathrm{NaOH}\left(\mathrm{aq}\right)\to \mathrm{Fe}{\left(\mathrm{OH}\right)}_3\downarrow +\mathrm{NaCl}\left(\mathrm{aq}\right)$

• In the above-unbalanced reaction, there is $1$ an atom for oxygen and hydrogen on the reactant side while there are $3$ atoms for oxygen and hydrogen on the product side.
• Thus, in order to balance this, one will write a stoichiometric coefficient $3$ in front of $\mathrm{NaOH}$ on reactant side, which balances oxygen and hydrogen atoms but the sodium atom in the product side gets unbalanced as shown below.
• ${\mathrm{FeCl}}_3\left(\mathrm{aq}\right)+3\mathrm{NaOH}\left(\mathrm{aq}\right)\to \mathrm{Fe}{\left(\mathrm{OH}\right)}_3\downarrow +\mathrm{NaCl}\left(\mathrm{aq}\right)$
• So, to balance this equation, one will write a stoichiometric coefficient $3$ in front of $\mathrm{NaCl}$ which will balance other atoms too.

Balanced chemical reaction:

${\mathrm{FeCl}}_3\left(\mathrm{aq}\right)+3\mathrm{NaOH}\left(\mathrm{aq}\right)\to \mathrm{Fe}{\left(\mathrm{OH}\right)}_3\downarrow +3\mathrm{NaCl}\left(\mathrm{aq}\right)$