Question

Calculate the work done (in Joules) when 43.3 g of Tin dissolves in excess acid at 1.05 atm and 23 degree Celsius. Assume ideal gas behavior.

Sn(s) + 2H⁺ (aq) -> Sn²⁺(aq) + H₂ (g)

Answer 1

Assuming ideal gas behavior, 1 mole of H2 gas at 21 C (294.15 K) and 1 atm (1.013*10^5 pascals) occupies a volume of:

V = n*R*T/P

so the work done per mole of Sn that reacts is:

work = 1mol*R*T

work = 1mol * 8.314J/mol*K * 296 K

work = 2.460*10^3 J

The atomic mass of Sn is 118.71 gm/mol, so 43.3 gm of Sn is 0.364 moles

The work done by the reaction of 60 gm of Sn under the specified conditions is therefore:

0.364 * 2.460*10^3 J = 0.895*10^3 J

Answer 2