What volume of $3.0 M$ $K O H$ would be required to react with the $S O_{2}$ produced if a concentration of $S O_{2}$ is $2 M$ and volume of $S O_{2}$ is $8.32$ L ?

2.77

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5.54

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1.38

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8.31

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Solution

The correct option is B $5.54$ L
The balanced reaction is as follows:
$2 K O H + S O_{2} \to K_{2} S O_{3} + H_{2} O$
Moles of $K O H = 2 \times$ moles of $S O_{2} = 2 \times 8.32$
$M \times V_{L} =$ Moles of $S O_{2}$
$3 \times V_{L} = 2 \times 8.32$
$V_{L} = 5.54 L$

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What volume of 3.0 M KOH would be required to react with the SO2 produced if a concentration of SO2 is 2 M and volume of SO2 is 8.32 L ?

The correct option is B 5.54 LThe balanced reaction is as follows:2KOH+SO2→K2SO3+H2OMoles of KOH=2× moles of SO2=2×8.32M×VL= Moles of SO23×VL=2×8.32VL=5.54 L

What volume of $3.0 M$ $K O H$ would be required to react with the $S O_{2}$ produced if a concentration of $S O_{2}$ is $2 M$ and volume of $S O_{2}$ is $8.32$ L ?

The correct option is B $5.54$ L
The balanced reaction is as follows:
$2 K O H + S O_{2} \to K_{2} S O_{3} + H_{2} O$
Moles of $K O H = 2 \times$ moles of $S O_{2} = 2 \times 8.32$
$M \times V_{L} =$ Moles of $S O_{2}$
$3 \times V_{L} = 2 \times 8.32$
$V_{L} = 5.54 L$