Question

A gas ‘A’ reacts with another gas ‘B’ in the presence of a catalyst to give a colourless gas ‘C’. The gas ‘C’ when comes in contact with air produces a brown gas ‘D’. The solution of ‘A’ in water turns red litmus blue. Explain the observations.

Answer 1

As the 'A' turns red litmus blue, it is a base. Now the gas 'A' combines with 'B' in the presence of a catalyst to give colourless gas Nitrogen monoxide. It reacts with oxygen to give brown gas, which is Nitrogen dioxide.

A = \[\ce{NH3}\]

B = \[\ce{O2}\]

C = \[\ce{NO}\]

D = \[\ce{NO2}\]

Catalyst = \[\ce{Pt}\]

Reactions:

With air \[\ce{NO}\] forms → \[\ce{NO2}\] [Brown gas] - gas 'D'

\[\ce{4NH3 + 5O2 ->[Pt 800°C] 4NO↑ + 6H2O + Heat}\]

\[\ce{2NO + O2 -> \underset{(Brown gas)}{2NO2}}\]

\[\ce{NH3}\] in water forms \[\ce{NH4OH}\] which turns red litmus blue.

\[\ce{NH3 + H2O -> NH4OH}\]