Calculate the work done when 40.0 g of tin dissolves in excess acid at 1.00 atm and 29°C. Assume ideal gas behavior.
Sn(s) + 2 H+(aq) → Sn2+(aq) + H2(g)
To calculate the work done when tin dissolves in excess acid, we first need to determine the number of moles of tin that reacted.
1. Convert the mass of tin to moles:
Number of moles = mass / molar mass
The molar mass of tin (Sn) is 118.71 g/mol.
Number of moles = 40.0 g / 118.71 g/mol = 0.3371 mol
Next, we need to determine the change in volume of the system. Since the equation shows that one mole of tin reacts to produce one mole of hydrogen gas, the change in volume is equal to the moles of tin reacted (0.3371 mol).
Now, we can calculate the work done using the following equation:
Work done = -P * ΔV
where P is the pressure and ΔV is the change in volume.
In this case, the pressure is given as 1.00 atm, and the change in volume is equal to 0.3371 mol.
So, the work done = - (1.00 atm) * (0.3371 mol) = -0.3371 L ∙ atm
Note that the negative sign indicates that work is done on the system, as the volume decreases.
Therefore, the work done when 40.0 g of tin dissolves in excess acid at 1.00 atm and 29°C is approximately -0.3371 L ∙ atm.