The volume of H ₂ obtained at STP , when Mg obtained by passing a current of 0.5 ~A through molten MgCl ₂ for 32.2 min is treated with excess of dilute HCl is approximately [Equivalent mass of Mg =12 ]

Step-by-step solution

I=0.5 ~A gathered t=32.2 60=1932.0 ~s Mg obtained =Z I t= 12 96500 0.5 1932.0=0.12012 ~g Mg +2 HCl MgCl ₂+ H ₂ 24 ~g of Mgg gives =22400 ~cm ³ of H ₂ 0.12012 ~g Mg gives = 22400 24 0.12012=112 ~cm ³ gathered