Answer :

The reaction of aluminium with caustic soda can be represented as:


At STP (273.15 K and 1 atm), 54 g (2 × 27 g) of Al gives 3 × 22400 ml of Hydrogen


0.15 g Al gives i.e. 3×22400×0.15/54 = 186.67 ml of Hydrogen.


At STP,


Let the volume of dihydrogen V2 be at p2 = 0.987 atm (since 1 bar = 0.987 atm) and T2 = 293.15K.


20°C = (273.15 + 20) K = 293.15 K.


= constant


So,



V2 = 203ml


Therefore, 203 ml of dihydrogen will be released.


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