Question

Consider the following series of reactions:
$I)C{l}_2+2NaOH\to NaCl+NaClO+H_{2}O$
$II)3NaClO\to 2NaCl+NaCl{O}_3$
$III)4NaCl{O}_3\to 3NaCl{O}_4+NaCl$
How much $C{l}_2$ is required to prepare 122.5 g of $NaCl{O}_4$ by the above sequential reactions?
(Assume yield of each reaction is 100 %)

• A. 284 g
• B. 213 g
• C. 142 g
• D. 71 g

Answer 1

The correct option is A 284 g

$C{l}_2+2NaOH\to NaCl+NaClO+H_{2}O$
$3NaClO\to 2NaCl+NaCl{O}_3$
$4NaCl{O}_3\to 3NaCl{O}_4+NaCl$
From the above series of reaction,
$1molofC{l}_2=1moleNaClO=\frac{1}{3}moleofNaCl{O}_3=\frac{3}{4}\times \frac{1}{3}molofNaCl{O}_4$
So, 1 mole of $C{l}_2=\frac{1}{4}moleofNaCl{O}_4$
122.5 g of $NaCl{O}_4=\frac{122.5}{122.5}=1mole$
To get 1 mole of $NaCl{O}_4$ , amount of $C{l}_2$ required is $=\frac{1}{\frac{1}{4}}=4mole$
Weight of $C{l}_2=4\times 71=284g$