
When one gram mole of $KMn{O_4}$ reacts with $HCl$, the volume of chlorine liberated at NTP will be:
A. 11.2 litres
B. 22.4 litres
C. 44.8 litres
D. 56.0 litres
Answer
549k+ views
Hint: Potassium permanganate, a strong oxidizing agent is used as a reactant for producing manganese compounds. When potassium permanganate ($KMn{O_4}$) reacts with concentrated hydrochloric acid ($HCl$), the products formed are manganese(II) chloride, potassium chloride, water and chlorine gas.
Formula used: As derived from ideal gas law,
$V = n \times 22.4L$
$n$= moles of the gas produced
$V = $ Volume of the gas at NTP
Complete step by step answer:
The chemical reaction between potassium permanganate and concentrated hydrochloric acid is shown below.
$2KMn{O_4} + 16HCl \to 2KCl + 2MnC{l_2} + 8{H_2}O + 5C{l_2}$
As per chemical equation above, 2 moles of potassium permanganate ($KMn{O_4}$) reacts with 16 moles of hydrochloric acid to produce 2 moles of potassium chloride, 2 moles of manganese chloride, 8 moles of water and 5 moles of chlorine gas were released.
Therefore, if 1 mole of $KMn{O_4}$ reacted with 8 moles of $HCl$, then the amount of chlorine gas that will liberate is $\dfrac{5}{2} = 2.5moles$.
At NTP, one mole of any gas will occupy 22.4L of volume.
Now, to calculate the volume of 2.5 moles of chlorine gas at NTP, we have to use the following formula.
$V = n \times 22.4L$
On substituting the values in the formula, we get,
$V = 2.5 \times 22.4L$
$\therefore V = 56L$
Therefore, the volume of chlorine gas at NTP will be 56L or 56 litres.
So, the correct answer is Option D.
Additional information:
The formula used for calculating the volume of chlorine gas at NTP is derived from the ideal gas law equation.
According to the ideal gas law,
$PV = nRT$
We know that, at NTP,
$P = 1atm$
$n = 1mole$
$R = 8.314J/mol.{K^{ - 1}}$
$T = 273K$
On putting these values in ideal gas law equation, we get
$V = 22.4L$
Note: Students might get confused about STP and NTP. There is a difference between STP and NTP. STP is Standard Temperature and Pressure and NTP is Normal Temperature and Pressure. As stated by IUPAC, at STP, temperature is considered as 0$^\circ $C and pressure is 1 bar. At NTP, pressure is 1 atm and temperature is 20$^\circ $C.
Formula used: As derived from ideal gas law,
$V = n \times 22.4L$
$n$= moles of the gas produced
$V = $ Volume of the gas at NTP
Complete step by step answer:
The chemical reaction between potassium permanganate and concentrated hydrochloric acid is shown below.
$2KMn{O_4} + 16HCl \to 2KCl + 2MnC{l_2} + 8{H_2}O + 5C{l_2}$
As per chemical equation above, 2 moles of potassium permanganate ($KMn{O_4}$) reacts with 16 moles of hydrochloric acid to produce 2 moles of potassium chloride, 2 moles of manganese chloride, 8 moles of water and 5 moles of chlorine gas were released.
Therefore, if 1 mole of $KMn{O_4}$ reacted with 8 moles of $HCl$, then the amount of chlorine gas that will liberate is $\dfrac{5}{2} = 2.5moles$.
At NTP, one mole of any gas will occupy 22.4L of volume.
Now, to calculate the volume of 2.5 moles of chlorine gas at NTP, we have to use the following formula.
$V = n \times 22.4L$
On substituting the values in the formula, we get,
$V = 2.5 \times 22.4L$
$\therefore V = 56L$
Therefore, the volume of chlorine gas at NTP will be 56L or 56 litres.
So, the correct answer is Option D.
Additional information:
The formula used for calculating the volume of chlorine gas at NTP is derived from the ideal gas law equation.
According to the ideal gas law,
$PV = nRT$
We know that, at NTP,
$P = 1atm$
$n = 1mole$
$R = 8.314J/mol.{K^{ - 1}}$
$T = 273K$
On putting these values in ideal gas law equation, we get
$V = 22.4L$
Note: Students might get confused about STP and NTP. There is a difference between STP and NTP. STP is Standard Temperature and Pressure and NTP is Normal Temperature and Pressure. As stated by IUPAC, at STP, temperature is considered as 0$^\circ $C and pressure is 1 bar. At NTP, pressure is 1 atm and temperature is 20$^\circ $C.
Recently Updated Pages
Why are manures considered better than fertilizers class 11 biology CBSE

Find the coordinates of the midpoint of the line segment class 11 maths CBSE

Distinguish between static friction limiting friction class 11 physics CBSE

The Chairman of the constituent Assembly was A Jawaharlal class 11 social science CBSE

The first National Commission on Labour NCL submitted class 11 social science CBSE

Number of all subshell of n + l 7 is A 4 B 5 C 6 D class 11 chemistry CBSE

Trending doubts
Differentiate between an exothermic and an endothermic class 11 chemistry CBSE

10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

State the laws of reflection of light

