The work done in an open vessel at 300 K, when 112 g iron reacts with dil. HCl is:
`(R = 2 "cal mol"^(-1) K^(-1))`

Options

A. 1200 cal
B. 600 cal
C. 300 cal
D. 200 cal

Answer

A

Explanation

`underset(112/56 = 2 "mol") underset("1 mol")(Fe) + 2HCl to FeCl_(2) + underset("2 mol")underset("1 mol")(H_(2))`
Now, `V_(2) = V_(H_(2))` and `V_(1) =0` (For state + aqueous)
Work done = `-rho_("ext") DeltaV`
`=-Delta(PV) =- n_(H_(2)) RT`
`=-2 xx 2 xx 300 = -1200` cals (Work of expansion)