Question

Sodium chlorate, ${\rm{NaCl}}{{\rm{O}}_{\rm{3}}}$, can be prepared by the following series of reactions.
${\rm{2KMn}}{{\rm{O}}_{\rm{4}}} + {\rm{16HCl}} \to {\rm{2KCl}} + {\rm{2MnC}}{{\rm{l}}_{\rm{2}}} + {\rm{8}}{{\rm{H}}_{\rm{2}}}{\rm{O}} + {\rm{5C}}{{\rm{l}}_{\rm{2}}}\\\ {\rm{6C}}{{\rm{l}}_{\rm{2}}} + {\rm{6Ca}}{\left( {{\rm{OH}}} \right)_{\rm{2}}} \to {\rm{Ca(Cl}}{{\rm{O}}_{\rm{3}}}{{\rm{)}}_{\rm{2}}} + {\rm{5CaC}}{{\rm{l}}_{\rm{2}}} + {\rm{6}}{{\rm{H}}_{\rm{2}}}{\rm{O}}\\\ {\rm{Ca(Cl}}{{\rm{O}}_{\rm{3}}}{{\rm{)}}_2} + {\rm{N}}{{\rm{a}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_4} \to {\rm{CaS}}{{\rm{O}}_{\rm{4}}} + {\rm{2NaCl}}{{\rm{O}}_{\rm{3}}}$
What mass of ${\rm{NaCl}}{{\rm{O}}_{\rm{3}}}$ can be prepared from ${\rm{100}}\;{\rm{ml}}$ of concentrated HCl (density ${\rm{1}}{\rm{.28 g/ml}}$ and $36.5\%$ by mass)?
Assume all other substances are present in excess amounts.
A. $14.2\;{\rm{g}}$
B. $23.3\;{\rm{g}}$
C. $34.7\;{\rm{g}}$
D. $45.8\;{\rm{g}}$

Answer 1

We know that the number of moles of any substituents can be determined with the help of mass along with the renowned quantity that is molar mass.

Complete step by step solution
Given, the density of concentrated HCl is ${\rm{1}}{\rm{.28 g/ml}}$.
The volume of concentration HCl is ${\rm{100}}\;{\rm{ml}}$.
The mass percentage of HCl is $36.5\%$.
The given reaction is shown below.
${\rm{2KMn}}{{\rm{O}}_{\rm{4}}} + {\rm{16HCl}} \to {\rm{2KCl}} + {\rm{2MnC}}{{\rm{l}}_{\rm{2}}} + {\rm{8}}{{\rm{H}}_{\rm{2}}}{\rm{O}} + {\rm{5C}}{{\rm{l}}_{\rm{2}}}\\\ {\rm{6C}}{{\rm{l}}_{\rm{2}}} + {\rm{6Ca}}{\left( {{\rm{OH}}} \right)_{\rm{2}}} \to {\rm{Ca(Cl}}{{\rm{O}}_{\rm{3}}}{{\rm{)}}_{\rm{2}}} + {\rm{5CaC}}{{\rm{l}}_{\rm{2}}} + {\rm{6}}{{\rm{H}}_{\rm{2}}}{\rm{O}}\\\ {\rm{Ca(Cl}}{{\rm{O}}_{\rm{3}}}{{\rm{)}}_2} + {\rm{N}}{{\rm{a}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_4} \to {\rm{CaS}}{{\rm{O}}_{\rm{4}}} + {\rm{2NaCl}}{{\rm{O}}_{\rm{3}}}$
The molar mass of HCl is ${\rm{36}}{\rm{.5}}\;{\rm{g/mol}}$.
The molar mass of ${\rm{NaCl}}{{\rm{O}}_{\rm{3}}}$ is $106.44\;{\rm{g/mol}}$.
The mass of HCl can be calculated by using the formula given below.
${\rm{M}} = {\rm{d}} \times {\rm{V}}$
Where, M is the mass of HCl, d is the density of HCl, and V is the volume of HCl.
Substitute all the respective values in the above equation.
${\rm{M}} = {\rm{1}}{\rm{.28 g/ml}} \times {\rm{100}}\;{\rm{ml}}\\\ = 128\;{\rm{g}}$
The mass percentage of HCl is given $36.5\%$ which means $36.5\% \times 128\;{\rm{g}} = 46.72\;{\rm{g}}$.
The number of moles of HCl can be calculated by using the formula given below.
${\rm{Number}}\;{\rm{of}}\;{\rm{moles}} = \dfrac{{{\rm{Mass}}}}{{{\rm{Molar}}\;{\rm{mass}}}}$
Substitute all the respective values in the above equation.
${\rm{Number}}\;{\rm{of}}\;{\rm{moles}} = \dfrac{{46.72\;{\rm{g}}}}{{{\rm{36}}{\rm{.5}}\;{\rm{g/mol}}}}\\\ = 1.28\;{\rm{mol}}$
According to the above chemical reaction, it is seen that $\dfrac{{96}}{5}$ moles of HCl gives $2$ moles of ${\rm{NaCl}}{{\rm{O}}_{\rm{3}}}$.
So, 1 mole of HCl gives $\dfrac{{2 \times 5}}{{96}}$ moles of ${\rm{NaCl}}{{\rm{O}}_{\rm{3}}}$ and $1.28$moles of HCl will give $\dfrac{{2 \times 5}}{{96}} \times 1.28\;{\rm{mol}} = 0.133\;{\rm{mol}} \\\$.
The mass of ${\rm{NaCl}}{{\rm{O}}_{\rm{3}}}$ can be calculated by using the formula given below.
{\rm{Mass}} = {\rm{Number}}\;{\rm{of}}\;{\rm{moles}} \times {\rm{Molar}}\;{\rm{mass}} \\\
Substitute all the respective values in the above equation.
${\rm{Mass}} = 0.133\;{\rm{mol}} \times 106.44\;{\rm{g/mol}}\\\ = 14.15\;{\rm{g}}\\\ \approx 14.2\;{\rm{g}}$

Hence, the correct option for this question is A that is $14.2\;{\rm{g}}$.

Note:
The mass, the volume and the density all comes under the category of physical chemistry.