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

Options

A. `-2.48` kJ
B. `-2.22` kJ
C. `+2.22` kJ
D. `+2.48` kJ

Answer

A

Explanation

`Fe+2HCl + FeCl_2 + H_2`
1 mole of Iron liberates 1 mole of hydrogen gas
55.85 g Iron = 1 mole Iron
`therefore` n=1
`T=25^@C`=298 K
`w=-PDeltaV`
`w=-P((nRT)/P)`
w=-nRT
w=-1 x 8.314 x 298 J
w=-2477.57 J
w=-2.48 kJ