Question

The work done in an open vessel when $113gm$ of $Fe$ reacts with dil. $HCl$ to form $Fe{Cl}_2$ at ${27}^{o}C$ will be:

• A. $w=-1200$ cal
• B. $w=-200$ cal
• C. $w=-300$ cal
• D. $w=-400$ cal

Answer 1

The correct option is A $w=-1200$ cal

The reaction involved is :
Fe(s) + 2 HCl(aq) →→ FeCl2 + H2
​Work done (W) = - P(V2 - V1)
Mass of Fe used for reaction = 113 g
Molar mass of Fe = 56 g/mol
Moles of Fe used =1135611356 moles = 2.018 moles
Thus moles of H2produced in the reaction are = ​2.018 moles
For the reaction V1 = 0 ( Initially no H2 is produced)
and V2 = nRT/P
Thus W = - P (V2 - 0) = - PV2
Substituting the value of V2
​W = −P×nRTP-P×nRTP
W= - nRT
Substituting, n= 2.018
R = 2cal
T = 270 C = 300 K
W = - 2.018 ××2 ××300 = 1210.8 cal~ 1200 cal.

Option A is correct answer