When one gram mole of $K M n O_{4}$ reacts with $H C l$ , the volume of chlorine liberated at NTP will be :

11.2 litres

No worries! We‘ve got your back. Try BYJU‘S free classes today!

22.4 litres

No worries! We‘ve got your back. Try BYJU‘S free classes today!

44.8 litres

No worries! We‘ve got your back. Try BYJU‘S free classes today!

56.0 litres

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D 56.0 litres
Potassium permanganate reacts with hydrochloric acid according the following equation:

$2 K M n O_{4} + 16 H C l \to 2 K C l + 2 M n C l_{2} + 8 H_{2} O + 5 C l_{2}$

2 moles of $K M n O_{4}$ produces 5 moles of $C l_{2}$

1 mole of $K M n O_{4}$ will produce 2.5 moles of $C l_{2}$

Volume of 2.5 moles of $C l_{2} = 2.5 \times 22.4 = 56 l i t r e s$

Suggest Corrections

Similar questions

K M n O_{4}

H C l

o n e g r a m m o l e

K M n O_{4}

N T P

11.2

What will be the volume (in litres) of oxygen liberated (at STP) when 17 grams of $H_{2}$ $O_{2}$ decomposed on heating?

H_{2} O

11.2

H_{2}

O_{2}

Join BYJU'S Learning Program