Question

Calculate the work done when $11.2g$ of iron is dissolved in $HCl$ at ${25}^{o}C$ in (i) a closed vessel and (ii) an open beaker when the atmospheric pressure in $1$ atm.

Answer 1

$\text{Iron reacts with HCl acid to produce dihydorgen gas as :}$

$Fe$+ $2HCl$------> $FeC{l}_2$+ $H_2$

Thus, 1 mole of Fe, i.e., 56 g Fe produces $H_2$ gas = 1 mol.
11.2 g Fe will $H_2$ produce gas = 1/56 x 11.2 = 0.2 mol

$\left(i\right)$ If the reaction is carried out in closed vessel, ∆V = 0
$W=-P_{ext}$ x ∆V = 0

$\left(ii\right)$ $\text{if the reaction is carried out in open beaker (external pressure being 1atm)(initial volume is = 0)}$
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