Question

A sample of chalk weighing $1.5g$ was dissolved in $200ml\;0.1M\;dil\;HCl$. The solution required $50ml$ of $0.2N\;NaOH$ to neutralize the excess acid. What is the weight of $CaC{O_3}$ in the sample?

Answer 1

Hint: We know that chalk contains calcium carbonate $\left( {CaC{O_3}} \right)$ with a small amount of silt and clay. When the calcium carbonate reacts with $HCl$ we get,
$CaC{O_3} + 2HCl \to CaC{l_2} + {H_2}C{O_3}$

Complete step by step solution:
We can see $HCl$ is titrated with $NaOH$ to neutralise the excess acid. Here, we can see that there is some amount of excess acid when $HCl$ was dissolved in the chalk. Which means that when chalk (which is the $CaC{O_3}$) there is some excess $HCl$ present in it because there is a lesser amount $CaC{O_3}$ is available.
We require an extra amount of $NaOH$ to neutralise this excess acid. So we can see that the excess amount of $HCl$ will equal to this extra amount of $NaOH$.
i.e.,$moles\;of\;excess\;HCl\; = \;moles\;of\;NaOH$
Now, to find the moles of $NaOH$, we have given that it has $50ml$ in volume and $0.2N$ in normality.
From the equation of normality,
$Normality\left( N \right) = \dfrac{{number\;of\;gram\;eq\;moles\;of\;solute}}{{Volume\;\left( L \right)}}$
Number of moles of $NaOH = N \times V$
$ \Rightarrow Number\;of\;moles\;NaOH\; = \;0.2 \times \dfrac{{50}}{{1000}}$(volume in L)
Now when we look into the number of moles of $HCl$ that reacted with $CaC{O_3}$will be equal to the difference between the number of moles excess $HCl$ taken and number of moles of $NaOH$
Number of moles of $HCl = N \times V$
Now, to find the moles of $HCl$, we have given that it has $200\,ml$ in volume and $0.1N$ in normality.
$ \Rightarrow Number\;of\;moles\;NaOH\; = \;0.1 \times \dfrac{{200}}{{1000}}$(volume in L)
Number of moles of HCl reacted with $CaC{O_3} = 0.1 \times \dfrac{{200}}{{1000}} - 0.2 \times \dfrac{{50}}{{1000}}$
$ = 0.02 - 0.01$
$ = 0.01$
Now from the reaction stoichiometry,
$CaC{O_3} + 2HCl \to CaC{l_2} + {H_2}C{O_3}$
2 moles of $HCl$is required to react with 1 mole of $CaC{O_3}$
i.e., 1 mole of $HCl$ is required to react with 0.5 moles of $CaC{O_3}$
i.e., 0.01 moles of $HCl$ is required to react with $0.5 \times 0.01$ mole of $CaC{O_3}$
Therefore, we can say that 0.005 moles of $CaC{O_3}$ is used.
Mass of 1 mole of $CaC{O_3}$is 100g
So, 0.005 moles of $CaC{O_3}$ has 0.5g.
Therefore, the percentage weight of $CaC{O_3}$ in sample $ = \dfrac{{mass\;of\;CaC{O_3}}}{{mass\;of\;sample}} \times 100$
Mass of sample is 1.5 g (Given)
The percentage weight of $CaC{O_3}$ in sample $ = \dfrac{{0.5}}{{1.5}} \times 100$
=33.3%

Percentage weight of $CaC{O_3}$ in the sample is 33.3%.

Note:
While using the reaction here we should not forget to balance the equation, otherwise, the solution will be wrong. In this reaction, we did use the balanced reaction.