Question

$KCl{O}_4$ is obtained from the following sequence of reactions:

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

The correct statement among the following is/are

• A. To produce $0.1mole$ of $KCl{O}_4$ , $8.96L$ of $C{l}_2$ at STP is required.
• B. If the yield of the each reaction is $80\%$, then moles of $KCl{O}_4$ produced from $1.2moles$ of $C{l}_2$ will be $0.1536$
• C. Total moles of $KCl$ formed from $1.2moles$ of $C{l}_2$ will be $2.1moles$ of $KCl$
• D. Total moles of $KCl$ formed from $53.76L$ of $C{l}_2$ at STP will be $4.2moles$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A To produce $0.1mole$ of $KCl{O}_4$ , $8.96L$ of $C{l}_2$ at STP is required.
B If the yield of the each reaction is $80\%$, then moles of $KCl{O}_4$ produced from $1.2moles$ of $C{l}_2$ will be $0.1536$
C Total moles of $KCl$ formed from $1.2moles$ of $C{l}_2$ will be $2.1moles$ of $KCl$
D Total moles of $KCl$ formed from $53.76L$ of $C{l}_2$ at STP will be $4.2moles$


$12C{l}_2+24KOH\to 21KCl+3KCl{O}_4+12H_{2}O$

A) From stoichiometry,

$12\times 22.4$ gives 3 moles of $KCl{O}_4$

Volume of $C{l}_2$ required to form 0.1 moles of $KCl{O}_4$ produced = $1\frac{12\times 22.4}{3}\times 0.1=8.96L$

B) $12C{l}_2+24KOH\to 12KCl+12KClO+12H_{2}O$
$12KClO\to 8KCl+4KCl{O}_3$
$4KCl{O}_3\to KCl+3KCl{O}_4$

12 moles of $C{l}_2$ gives 12 moles of $KClO$
moles of $KClO$ that can be obtained from 1.2 moles of $C{l}_2$ = 1.2
Yield of the reaction is 80 %
Hence moles that can be obtained $=0.8\times 1.2=0.96$

12 moles of $KClO$ gives 4 moles of $KCl{O}_3$
moles of $KCl{O}_3$ that can be obtained from 0.96 moles of KClO $=\frac{4\times 0.96}{12}=0.32$
Yield of the reaction is 80 %.
Hence the actual number of moles $0.8\times 0.32=0.256$

4 moles of $KCl{O}_3$ gives 3 moles of $KCl{O}_4$
Hence moles of $KCl{O}_4$ that can be obtained from 0.256 moles of $KCl{O}_3$ = $\frac{3\times 0.256}{4}=0.192$
Yield of the reaction is 80%.
Hence actual moles of $KCl{O}_4$ that can be obtained$=0.192\times 0.8=0.1536$


C) 12 moles of $C{l}_2$ gives 21 moles of $KCl$ and hence 1.2 moles gives 2.1 moles of $KCl$
D) volume of $C{l}_2$ required to form 4.2 moles of $KCl$ produced = $1\frac{12\times 22.4}{21}\times 4.2=53.76L$