Calculate the work done when 60.0 g of tin dissolves in excess acid at 1.00 atm and 20°C. Assume ideal gas behavior.

The work done is different from the heat released, and can be calculated as the pressure multiplied by the volume change. The volume change is the result of the liberation of H2 gas by the reaction.

60.0 g of Sn is 0.505 moles.

In a reaction between Sn and an acid, two possible products can result,corresponding to a stannic or stannous salt. If a stannic product forms, as in the reaction
Sn + 4 HCl -> 2 H2 + SnCl4,
then 1.01 moles of H2 will be formed. The gas volume created at these NTP conditions will be
delta V = 24.1 liters = 24.1*10^-3 m^3.

The work done is P*deltaV Joules, with P =1.013*10^5 Pascals.

To calculate the work done when tin dissolves in excess acid, we need to use the equation for work done in an ideal gas system. The equation is:

Work = -PΔV

Where:
- Work is the work done in Joules (J)
- P is the pressure in atmospheres (atm)
- ΔV is the change in volume in liters (L)

First, we need to find the change in volume when 60.0 g of tin dissolves. To do this, we need to know the molar mass of tin (Sn), which is 118.71 g/mol.

Next, we convert the mass of tin to moles by dividing the given mass by the molar mass.
moles of tin = mass of tin / molar mass of tin
moles of tin = 60.0 g / 118.71 g/mol

Once we have the moles of tin, we need to use the balanced chemical equation for the reaction between tin and acid to determine the change in moles of gas. Let's assume the balanced equation is:

Sn + 2HCl -> SnCl2 + H2

From the balanced equation, we can see that for every mole of tin reacted, one mole of hydrogen gas (H2) is produced. Therefore, the change in moles of gas is equal to the number of moles of tin.

Change in moles of gas = moles of tin

Now, we can calculate the change in volume using the ideal gas law equation:

PV = nRT

Where:
- P is the pressure in atmospheres (atm)
- V is the volume in liters (L)
- n is the number of moles of gas
- R is the ideal gas constant (0.0821 L.atm/mol.K)
- T is the temperature in Kelvin (K)

Before we calculate the change in volume, we need to convert the given temperature from Celsius to Kelvin.
T(K) = T(C) + 273.15
T(K) = 20°C + 273.15

Now we can substitute the values and solve for the change in volume:

(P1)(V1) = (n)(R)(T1)
(1.00 atm)(V1) = (moles of tin)(0.0821 L.atm/mol.K)(20°C + 273.15)

Solve for V1 to get the initial volume.

Next, we need to determine the final volume. Since the problem states that the tin dissolves in excess acid, we can assume that the volume does not change. Therefore, the change in volume (ΔV) is equal to zero (ΔV = 0).

Finally, we substitute the values into the equation for work done:

Work = -PΔV
Work = - (1.00 atm)(0 L)
Work = 0 J

Therefore, the work done when 60.0 g of tin dissolves in excess acid is 0 Joules.