Calculates the work done (in joules) when 42.8 g of tin dissolves in excess acid at 1.10 atm and 23 degrees Celcius. Assume ideal gas behaviour.
Try this.
Sn + 2HCl ==> SnCl2 + H2
mols Sn = 42.8/atomic mass Sn
mols H2 = mols Sn
Then PV = nRT
You know n, R, T, solve for PV and that is work of expansion of the gas. The unit is L*atm. Multiply by 101.325 to convert to joules.
To calculate the work done when tin dissolves in excess acid, we need to consider the ideal gas behavior of the produced gas.
The first step is to calculate the number of moles of tin. We can do this using the molar mass of tin, which is 118.71 g/mol.
Given the mass of tin is 42.8 g, the number of moles of tin can be calculated as follows:
moles of tin = mass of tin / molar mass of tin
= 42.8 g / 118.71 g/mol
≈ 0.360 moles
Since the reaction takes place at constant pressure (1.10 atm), the work done (w) can be calculated using the equation:
w = -PΔV
where P is the pressure and ΔV is the change in volume.
To find the change in volume, we can use the ideal gas law:
PV = nRT
where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature.
Given:
Pressure (P) = 1.10 atm
Temperature (T) = 23 °C = 23 + 273.15 = 296.15 K (Converted to Kelvin)
Gas constant (R) = 0.0821 L·atm/(mol·K) (ideal gas constant)
Rearranging the ideal gas law equation, we can solve for V:
V = (nRT) / P
Substituting the given values, we get:
V = (0.360 mol * 0.0821 L·atm/(mol·K) * 296.15 K) / 1.10 atm
Finally, we can calculate the work done using the formula:
w = -PΔV
Substituting the values for pressure (P) and change in volume (ΔV), we have:
w = -(1.10 atm) * [(0.360 mol * 0.0821 L·atm/(mol·K) * 296.15 K) / 1.10 atm]
Calculating this expression will give us the work done in joules.
To calculate the work done when 42.8 g of tin dissolves in excess acid, we need to consider the ideal gas behavior and use the ideal gas law.
The ideal gas law is given as:
PV = nRT
Where:
P = pressure in atmospheres (atm)
V = volume in liters (L)
n = number of moles
R = ideal gas constant (0.0821 L·atm/mol·K)
T = temperature in Kelvin (K)
First, we need to convert the given mass of tin (42.8 g) into moles. To do this, we divide the mass by the molar mass of tin.
The molar mass of tin (Sn) is approximately 118.7 g/mol.
42.8 g Sn / 118.7 g/mol = 0.360 mol Sn
Now, we can calculate the work done using the ideal gas law.
Since the problem states that there is excess acid, we can assume that the pressure in the system remains constant at 1.10 atm.
We also need to convert the temperature from Celsius to Kelvin. To do this, we add 273.15 to the temperature in Celsius.
Temperature in Kelvin = 23 °C + 273.15 = 296.15 K
Now we can substitute our values into the ideal gas law equation:
PV = nRT
(1.10 atm) ∙ V = (0.360 mol) ∙ (0.0821 L·atm/mol·K) ∙ (296.15 K)
We need to solve for the volume (V), so we rearrange the equation and substitute the known values:
V = (0.360 mol) ∙ (0.0821 L·atm/mol·K) ∙ (296.15 K) / (1.10 atm)
V = 6.814 L
Now that we have the volume, we can calculate the work done:
Work = Pressure x Change in Volume
The change in volume (ΔV) is equal to the final volume (V) minus the initial volume. Since we assume that the reaction occurs at constant pressure, the final volume is the same as the initial volume.
Therefore, the work done is:
Work = (1.10 atm) ∙ (6.814 L - 0 L) = 7.4954 atm·L
To convert from atm·L to joules, we need to use the conversion factor:
1 atm·L = 101.3 J
Therefore, the work done in joules is:
Work = 7.4954 atm·L × 101.3 J/atm·L = 759 J (rounded to three significant figures)
So, the work done (in joules) when 42.8 g of tin dissolves in excess acid at 1.10 atm and 23 degrees Celsius is approximately 759 J.