Question

$C{l}_2+2KOH\to KCl+KClO+H_{2}O$
$3KClO\to 2KCl+KCl{O}_3$
$4KCl{O}_3\to KCl+3KCl{O}_4$

2 moles of chlorine gas reacted with an excess of potassium hydroxide and 50 g of potassium perchlorate was obtained. Calculate the percentage yield of perchlorate in the reaction.
Molar mass of potassium perchlorate = 138.5 g/mol

• A. 72 %
• B. 60 %
• C. 90 %
• D. 49 %

Answer 1

The correct option is A 72 %

The complete balanced equation is:
$12C{l}_2+24KOH\to 21KCl+3KCl{O}_4+12H_{2}O$

12 moles of chlorine gives 3 moles of perchlorate.
2 moles of chlorine gives 0.5 moles of perchlorate.
Theoretical yield of perchlorate = 0.5 mole = 69.25 g
Experimental yield of perchlorate = 50 g
Percentage yield of perchlorate = $\frac{\text{experimental yield}}{\text{theoretical yield}}\times 100$ = $\frac{50\times 100}{69.25}=72$%