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Balancing Chemical Equations by Partial Equation Method

KClO 4 is obt...

Question

$K C l O_{4}$ is obtained from the following sequence of reactions:

$C l_{2} + 2 K O H \to K C l + K C l O + H_{2} O$
$3 K C l O \to 2 K C l + K C l O_{3}$
$4 K C l O_{3} \to K C l + 3 K C l O_{4}$ .

The correct statement among the following is/are

A

To produce

0.1 m o l e

of

K C l O_{4}

,

8.96 L

of

C l_{2}

at STP is required.

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B

If the yield of the each reaction is

80 %

, then moles of

K C l O_{4}

produced from

1.2 m o l e s

of

C l_{2}

will be

0.1536

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C

Total moles of

K C l

formed from

1.2 m o l e s

of

C l_{2}

will be

2.1 m o l e s

of

K C l

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D

Total moles of

K C l

formed from

53.76 L

of

C l_{2}

at STP will be

4.2 m o l e s

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Solution

The correct options are
A To produce $0.1 m o l e$ of $K C l O_{4}$ , $8.96 L$ of $C l_{2}$ at STP is required.
B If the yield of the each reaction is $80 %$ , then moles of $K C l O_{4}$ produced from $1.2 m o l e s$ of $C l_{2}$ will be $0.1536$
C Total moles of $K C l$ formed from $1.2 m o l e s$ of $C l_{2}$ will be $2.1 m o l e s$ of $K C l$
D Total moles of $K C l$ formed from $53.76 L$ of $C l_{2}$ at STP will be $4.2 m o l e s$

$12 C l_{2} + 24 K O H \to 21 K C l + 3 K C l O_{4} + 12 H_{2} O$

A) From stoichiometry,

$12 \times 22.4$ gives 3 moles of $K C l O_{4}$

Volume of $C l_{2}$ required to form 0.1 moles of $K C l O_{4}$ produced = $1 \frac{12 \times 22.4}{3} \times 0.1 = 8.96 L$

B) $12 C l_{2} + 24 K O H \to 12 K C l + 12 K C l O + 12 H_{2} O$
$12 K C l O \to 8 K C l + 4 K C l O_{3}$
$4 K C l O_{3} \to K C l + 3 K C l O_{4}$

12 moles of $C l_{2}$ gives 12 moles of $K C l O$
moles of $K C l O$ 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 $K C l O$ gives 4 moles of $K C l O_{3}$
moles of $K C l O_{3}$ 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 $K C l O_{3}$ gives 3 moles of $K C l O_{4}$
Hence moles of $K C l O_{4}$ that can be obtained from 0.256 moles of $K C l O_{3}$ = $\frac{3 \times 0.256}{4} = 0.192$
Yield of the reaction is 80%.
Hence actual moles of $K C l O_{4}$ that can be obtained $= 0.192 \times 0.8 = 0.1536$

C) 12 moles of $C l_{2}$ gives 21 moles of $K C l$ and hence 1.2 moles gives 2.1 moles of $K C l$
D) volume of $C l_{2}$ required to form 4.2 moles of $K C l$ produced = $1 \frac{12 \times 22.4}{21} \times 4.2 = 53.76 L$

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Similar questions

K C l O_{4}

is obtained from the following sequence of reactions:

C l_{2} + 2 K O H \to K C l + K C l O + H_{2} O

3 K C l O \to 2 K C l + K C l O_{3}

4 K C l O_{3} \to K C l + 3 K C l O_{4}

.

The correct statement among the following is/are

Q. From the following reaction sequences, the mass of

{C l}_{2}

needed to produce

100

K C l O_{4}

is (as the nearest integer) :

{C l}_{2} + 2 K O H \to K C l + K C l O + H_{2} O . . . (i)

3 K C l O \to 2 K C l + K C l O_{3} . . . (i i)

4 K C l O_{3} \to 3 K C l O_{4} + K C l . . . (i i i)

From the following reaction sequence
$C l_{2}$ + 2KOH $\to$ KCl + KClO + $H_{2} O$
3KClO $\to$ 2KCl + $K C l O_{3}$
$4 K C l O_{3}$ $\to$ $3 K C l O_{4} + K C l$

Calculate the mass of chlorine needed to produce 100g of $K C l O_{4}$

From the following reaction sequence
$C l_{2}$ + 2KOH $\to$ KCl + KClO + $H_{2} O$
3KClO $\to$ 2KCl + $K C l O_{3}$
$4 K C l O_{3}$ $\to$ $3 K C l O_{4} + K C l$

Calculate the mass of chlorine needed to produce 23g of $K C l O_{4}$

C l_{2} + 2 K O H \to K C l + K C l O + H_{2} O

3 K C l O \to 2 K C l + K C l O_{3}

4 K C l O_{3} \to K C l + 3 K C l 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

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