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)
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
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