What will happen when dilute sulphuric acid is added to sodium sulphite?

N a_{2} S O_{3} + H_{2} S O_{4} \to N a_{2} S O_{4} + S O_{3} + H_{2} O

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

N a_{2} S O_{3} + H_{2} S O_{4} \to N a_{2} S O_{3} + S O_{2} + H_{2} O

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

N a_{2} S O_{3} + H_{2} S O_{4} \to N a_{2} S O_{4} + S O_{2} + H_{2} O

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

N a_{2} S O_{4} + H_{2} S O_{4} \to N a_{2} S O_{4} + S O_{2} + H_{2} O

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

Open in App

Solution

The correct option is B $N a_{2} S O_{3} + H_{2} S O_{4} \to N a_{2} S O_{4} + S O_{2} + H_{2} O$
The reaction between dil. sulphuric acid and sodium sulphite occurs in two steps and is a double replacement reaction.

In the first step, sulphuric acid reacts with sodium sulphite to give sodium sulphate and sulphurous acid.
$H_{2} S O_{4} + N a_{2} S O_{3} \to N a_{2} S O_{4} + H_{2} S O_{3}$

In the second step, sulphurous acid, which is highly unstable decomposes spontaneously into sulphur dioxide and water.
$H_{2} S O_{3} \to S O_{2} + H_{2} O$

Thus, the overall reaction is written as:
$H_{2} S O_{4} + N a_{2} S O_{3} \to N a_{2} S O_{4} + S O_{2} + H_{2} O$

Suggest Corrections

Similar questions

NaOH+H2So4=Na2So4+H2o

N a_{2} C O_{3} + H_{2} S O_{4} \to

Complete the following reaction: $N a_{2} C O_{3} + H_{2} S O_{4} \to$ ___?

N a_{2} S O_{3} + 4 C l_{2} + 5 H_{2} O \to N a_{2} S O_{4} + H_{2} S O_{4} + 8 H C l

N a_{2} S O_{3}

N a_{2} S O_{3}

NaOH + H_{2} {SO}_{4} \to {Na}_{2} {SO}_{4} + H_{2} O

Join BYJU'S Learning Program