Question

Sodium chlorate, $NaCl{O}_3$, can be prepared by the following series of reactions.

$2KMn{O}_4+16HCl\to 2KCl+2MnC{l}_2+8H_{2}O+5C{l}_2$
$6C{l}_2+6Ca{\left(OH\right)}_2\to Ca{\left(Cl{O}_3\right)}_2+5CaC{l}_2+6H_{2O}$
$Ca{\left(Cl{O}_3\right)}_2+N{a}_{2}S{O}_4\to CaS{O}_4+2NaCl{O}_3$

What mass of $NaCl{O}_3$ can be prepared from $100$ ml of concentrated HCl (density $1.28$ g/ml and $36.5$% by mass)?

Assume all other substances are present in excess amounts.

• A. 14.2 g
• B. 23.3 g
• C. 34.7 g
• D. 45.8 g

Answer 1

The correct option is A 14.2 g

The molar masses of $HCl$ and $NaCl{O}_3$ are $36.5$ g/mol and $106.44$ g/mol respectively.
The number of moles present in $100$ ml of HCl is $100ml\times 1.28g/ml\times \frac{36.5}{100}\times \frac{1}{36.5}=1.28moles$
From the given chemical equations, it can be seen that $\frac{16\times 6}{5}$ moles of HCl will give $2$ moles of $NaCl{O}_3$
Hence, the mass of $NaCl{O}_3$ produced is $1.28\times \frac{1}{8}\times \frac{5}{6}\times 106.44=14.2g$