
When potassium bromide is treated with concentrated sulfuric acid reddish-brown gas evolved, gas is
A. Mixture of bromine and HBr
B. HBr
C. Bromine
D. None of these
Answer
161.7k+ views
Hint: Bromine is a halogen gas which is brown in colour. Sulfuric acid, a strong oxidising agent, oxidises potassium bromide to form a halogen.
Complete Step by Step Answer:
Potassium bromide, a salt of bromine reacts with hot and concentrated sulfuric acid, a strong acid in presence of manganese oxide to form potassium bisulphate or $KHSO_4$ , a sulphate of manganese that is manganese sulphate along with water and a reddish brown coloured gas that is bromine gas as products. The reaction can be given as follows-
$2KBr+MnO_2+3H_2SO_4\rightarrow MnSO_4+2KHSO_4+Br_2+2H_2O$
Thus in this reaction two moles of Potassium bromide, a salt of bromine reacts with three moles of hot and concentrated sulfuric acid, a strong acid in presence of one mole of manganese oxide to form two moles of potassium bisulphate or $KHSO_4$ , one mole of a sulphate of manganese that is manganese sulphate along with two moles of water and one mole of a reddish brown coloured gas that is bromine gas as products.
Thus we can write that when potassium bromide is treated with concentrated sulfuric acid reddish-brown gas evolved, the gas is bromine.
Thus the correct option is C.
Note: Bromine is the halogen gas. It is generally a brown coloured liquid. The other halogens like chlorine, fluorine and iodine are gas.
Complete Step by Step Answer:
Potassium bromide, a salt of bromine reacts with hot and concentrated sulfuric acid, a strong acid in presence of manganese oxide to form potassium bisulphate or $KHSO_4$ , a sulphate of manganese that is manganese sulphate along with water and a reddish brown coloured gas that is bromine gas as products. The reaction can be given as follows-
$2KBr+MnO_2+3H_2SO_4\rightarrow MnSO_4+2KHSO_4+Br_2+2H_2O$
Thus in this reaction two moles of Potassium bromide, a salt of bromine reacts with three moles of hot and concentrated sulfuric acid, a strong acid in presence of one mole of manganese oxide to form two moles of potassium bisulphate or $KHSO_4$ , one mole of a sulphate of manganese that is manganese sulphate along with two moles of water and one mole of a reddish brown coloured gas that is bromine gas as products.
Thus we can write that when potassium bromide is treated with concentrated sulfuric acid reddish-brown gas evolved, the gas is bromine.
Thus the correct option is C.
Note: Bromine is the halogen gas. It is generally a brown coloured liquid. The other halogens like chlorine, fluorine and iodine are gas.
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