The work done by the gas liberated when 50 g of iron (molar mass 55.85 g `mol^(-1)`) reacts with hydrochloric acid in an open beaker at `25^(@)C`

Options

A. Zero
B. `-2.2` kJ
C. 2.2kJ
D. 0.22kJ

Answer

B

Explanation

`underset(((50)/(55.85)"moles"))(Fe(s)) + 2HCl(aq) rarr FeCl_(2)(aq) + underset(((50)/(55.85) "moles"))(H_(2)(g))`
Moles of iron = Moles of hydrogen gas `= (50)/(55.85)= Delta n_(g)`
`R= 8.314 J "mole"^(-1) K^(-1)`
`T= 25+273= 298K`
Work done by `H_(2)` gas `W = - Delta n_(g)RT= (-50)/(55.85) xx 8.314 xx 298 xx 10^(-3)kJ= -2.2kJ`