Question

$C{l}_2+2NaOH\to NaCl+NaClO+H_{2}O$

$3NaClO\to 2NaCl+NaCl{O}_3$

$4NaCl{O}_3\to 3NaCl{O}_4+NaCl$

Consider the given series of reactions and calculate how much $C{l}_2$ is needed to prepare $106.5$ g of $NaCl{O}_3$?

• A. $284.0$ g
• B. $210.0$ g
• C. $142.0$ g
• D. $71.0$ g

Answer 1

The correct option is B $210.0$ g

1 mol $NaCl{O}_3$ = $23+35.5+3\times 16=106.5g$

As to prepare only 106.5 g, so it means only 1 mol is to prepare.

$3NaClO\to 2NaCl+NaCl{O}_3$

For preparing 1 mol of the $NaCl{O}_3$ , 3 mol of NaClO is required.

As per following reaction to prepare 3 mol of NaClO requires 3 mol of $C{l}_2$, as follows:

$3C{l}_2+6NaOH\to 3NaCl+3NaClO+3H_{2}O$.

Hence 3 mol of $C{l}_2$ is required.
Now weight of 3 mol $C{l}_2$ = $70\times 3=210g$

Hence, the correct option is $\left(B\right)$.