A 8.50-g sample of solid AlCl3·6H2O was heated such that the water turned to steam and was driven off. Assuming ideal behavior, what volume would that steam occupy at 1.00 atm and 100.0 °C?
AlCl3.6H2O ==> AlCl3 + 6H2O
mols AlCl3 = 8.5g/molar mass AlCl3 = approx 0.06 but you need a more accurate number.
Convert mols AlCl3 mols H2O using the coefficients in the balanced equation. That's about 6 x 0.06 = about 0.36.
Then plug that into for n in PV = nRT and solve for V in liters. Don't forget T must be in kelvin.
To calculate the volume of steam produced, you can use the ideal gas law equation:
PV = nRT
Where:
P = pressure (1.00 atm)
V = volume (unknown)
n = number of moles of gas
R = ideal gas constant (0.0821 L·atm/mol·K)
T = temperature in Kelvin (100.0 °C + 273.15)
First, we need to calculate the number of moles of water vapor produced. To do this, we'll determine the number of moles in the water initially present in the compound AlCl3·6H2O.
1 mole of AlCl3·6H2O contains 6 moles of H2O.
Therefore, the number of moles of H2O is:
8.50 g * (1 mol AlCl3·6H2O / 375.13 g AlCl3·6H2O) * 6 mol H2O / 1 mol AlCl3·6H2O
Now, let's calculate the number of moles of H2O:
8.50 g * 6 mol H2O / 375.13 g AlCl3·6H2O ≈ 0.136 mol H2O
Substituting the values into the ideal gas law equation, we have:
(1.00 atm) * V = (0.136 mol) * (0.0821 L·atm/mol·K) * (100.0 + 273.15) K
Simplifying:
V = (0.136 mol) * (0.0821 L·atm/mol·K) * (100.0 + 273.15) K / (1.00 atm)
Calculating the result will give us the volume of the steam produced at 1.00 atm and 100.0 °C.
To calculate the volume of steam at a given temperature and pressure, we need to use the ideal gas law equation:
PV = nRT
Where:
P: Pressure (in atm)
V: Volume (in liters)
n: Number of moles
R: Ideal gas constant (0.0821 L·atm/mol·K)
T: Temperature (in Kelvin)
First, we need to calculate the number of moles of steam. To do that, we use the molar mass of water, which is 18.015 g/mol.
Number of moles = mass / molar mass
Number of moles = 8.50 g / 18.015 g/mol
Number of moles ≈ 0.4713 mol
Next, we need to convert the temperature from Celsius to Kelvin. The Kelvin scale is obtained by adding 273.15 to the Celsius temperature.
Temperature in Kelvin = 100.0 °C + 273.15
Temperature in Kelvin = 373.15 K
Now, we can use the ideal gas law equation to calculate the volume:
PV = nRT
V = (nRT) / P
V = (0.4713 mol * 0.0821 L·atm/mol·K * 373.15 K) / 1.00 atm
V ≈ 15.53 L
Therefore, the volume of the steam would be approximately 15.53 liters at 1.00 atm and 100.0 °C.