44.8 L of chlorine gas at STP reacts with 280 g of KOH according to the following sequential reaction. Calculate the number of grams of $K C l O_{4}$ . Molar mass of $K C l O_{4} = 138.5$ $g m o l^{- 1}$
Molar mass of $K O H = 56$ $g m o l^{- 1}$

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

69.25

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138.5

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Solution

The correct option is A
69.25
22.4 L of chlorine gas at STP = 1 mole of chlorine gas
Given: 44.8 L of chlorine gas at STP
Number of moles of $C l_{2}$ = 2
Number of moles of KOH = $\frac{280}{56}$ = 5
Hence $C l_{2}$ is the limiting reagent.

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

3 moles $K C l O_{4}$ is formed from 12 moles of $C l_{2}$ and hence number of moles of $K C l O_{4}$ formed from 2 moles of $C l_{2} = \frac{3}{12} \times 2 = 0.5$
number of grams of $K C l O_{4} = 0.5 \times 138.5 = 69.25$ g

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Similar questions

Q. From the following reaction sequence:

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

Calculate the mass of chlorine that reacts with an excess of

K O H

needed to produce

100 g

K C l O_{4}

Q. From the following reaction sequence:

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

Calculate the mass of chlorine that reacts with an excess of

K O H

needed to produce

100 g

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 $K C l O_{4}$

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

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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44.8 L of chlorine gas at STP reacts with 280 g of KOH according to the following sequential reaction. Calculate the number of grams of KClO4. Molar mass of KClO4=138.5 gmol−1 Molar mass of KOH=56 gmol−1 Cl2+2KOH→KCl+KClO+H2O 3KClO→2KCl+KClO3 4KClO3→KCl+3KClO4