Question

Work done when 50g iron is dissolved in hcl at 27 degree celsius in an open beaker when atm pressure is 1 atm

Answer 1

Iron reacts with HCI acid to produce ${ H }{ 2 }$ gas as
Fe+ 2Hcl ------> ${ Fecl }{ 2 }$ + ${ H }_{ 2 }$

Thus, 1 mole of Fe, i.e., 56 g Fe produces ${ H }{ 2 }$ gas = 1 mol.
11.2 g Fe will produce ${ H }{ 2 }$ gas = 1/56 x 11.2 = 0.2 mol
(i) If the reaction is carried out in closed vessel, ∆V = 0
W = ${ -p }{ ext }$ ∆V = 0
(ii) If the reaction is carried out in open beaker (external pressure being 1 atm)
Initial volume = 0 (because no gas is present)
Final volume occupied by 0.2 mole of ${ H }{ 2 }$ at 25°C and 1 atm pressure can be calculated as follows pV = nRT
V = nrt/p
= 0.2 mol x 0.0821 L atm/K/mol x 298K / 1 atm
=4.89 L