Calculate the work done in an open vessel at `300K,` when `112g` ion reacts with dil. `HCl`.

Explanation

`Fe+2HCl rarr FeCl_(2)+H_(2)`
Work done `=-Pxx(V_(2)-V_(1))`
Since, mole of `Fe` used `=112/56=2`
`:.` Mole of `H_(2)` formed `=2`
Work is done by giving out `2`moles of `H_(2)`
`=-Pxx[V_(H_(2))-V_(i)] i.e., =-PxxV_(H_(2))`
[`V_(i)=0` for initial condition]
For `H_(2) PxxV_(H_(2))=nRT`
`:. V_(H_(2))=(nRT)/P`
`:.` Work done `=-Pxx(nRT)/P=-nRT`
`=-2xx2xx300=-1200 cal`