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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