Question

Excess of NaOH (aq) was added to $100mL$ of $FeC{l_2}$ (aq) resulting in the formation of $2.14g$ of $Fe{\left( {OH} \right)_3}$. What is the molarity of $FeC{l_3}$ ? Given molar mass of $Fe = 56gmo{l^{ - 1}}$ and molar mass of $Cl = 35.5gmo{l^{ - 1}}$
A) $0.6M$
B) $1.8M$
C) $0.3M$
D) $0.2M$

Answer 1

Write down the equation for the formation of $Fe{\left( {OH} \right)_3}$ from the reaction between Sodium hydroxide and Ferric chloride. Calculate the number of moles in $2.14g$ of $Fe{\left( {OH} \right)_3}$. From this value we can calculate the number of moles of $FeC{l_3}$ present and hence its molarity.

Complete step by step solution:
In order to find the molarity of Ferric chloride, let us write the equation for the reaction between Ferric chloride and Sodium hydroxide.
This reaction can be written as $NaOH + FeC{l_3}\left( {aq} \right) \to NaCl\left( {aq} \right) + Fe{\left( {OH} \right)_3}\left( {aq} \right)$
We have to balance this equation. The balanced equation can be written as $3NaOH + FeC{l_3}\left( {aq} \right) \to 3NaCl\left( {aq} \right) + Fe{\left( {OH} \right)_3}\left( {aq} \right)$

Now, we can see that in order to produce one mole of $Fe{\left( {OH} \right)_3}$ one mole of $FeC{l_3}$ is being utilized. Hence the ratio of number of moles of $Fe{\left( {OH} \right)_3}$ and $FeC{l_3}$ is $1:1$
So if we find the number of moles of $Fe{\left( {OH} \right)_3}$ produced, we can find the number of moles of $FeC{l_3}$ used.
Given to us the weight of $Fe{\left( {OH} \right)_3}$ as $2.14g$
We have to now calculate the molecular mass of $Fe{\left( {OH} \right)_3}$
The molecular mass of $Fe{\left( {OH} \right)_3}$ is $56 + 3\left( {17} \right) = 107gm/mol$
Number of moles of $Fe{\left( {OH} \right)_3}$ is $\dfrac{{2.14}}{{107}} = 0.02$
Since the ratio of moles of $Fe{\left( {OH} \right)_3}$ and $FeC{l_3}$ is $1:1$ the number of moles of $FeC{l_3}$ would be $0.02$

Now, $Molarity = Moles \times \dfrac{{1000}}{V}$
By substituting the acquired values in the above formula, we get the molarity of $FeC{l_3}$ as $0.02 \times \dfrac{{1000}}{{100}} = 0.02 \times 10 = 0.2$

Hence, the molarity of $FeC{l_3}$ is $0.2M$

Note: It is to be noted that balancing an equation is an important and required step before calculating the number of moles and moles ratio from it. One should also note that the units of certain quantities must be mentioned at the end.