A 2.70 g sample of solid Ni(CN)2⋅4H2O 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?
To calculate the volume of the steam, we can use the ideal gas law equation, which states:
PV = nRT
Where:
P = Pressure in atmospheres (1.00 atm)
V = Volume in liters (unknown)
n = Number of moles of gas
R = Ideal Gas Constant (0.0821 L·atm/(mol·K))
T = Temperature in Kelvin (100.0 °C = 373.15 K)
First, we need to determine the number of moles of water vapor produced from the given mass of Ni(CN)2⋅4H2O.
Step 1: Calculate the molar mass of Ni(CN)2⋅4H2O.
The molar mass of Ni(CN)2 is:
Ni: 1 atom x 58.69 g/mol = 58.69 g/mol
C: 2 atoms x 12.01 g/mol = 24.02 g/mol
N: 2 atoms x 14.01 g/mol = 28.02 g/mol
The molar mass of H2O is:
H: 8 atoms x 1.01 g/mol = 8.08 g/mol
O: 4 atoms x 16.00 g/mol = 64.00 g/mol
Adding them up:
Molar mass of Ni(CN)2⋅4H2O = 58.69 g/mol + 24.02 g/mol + 28.02 g/mol + 8.08 g/mol + 64.00 g/mol = 182.81 g/mol
Step 2: Calculate the number of moles of Ni(CN)2⋅4H2O.
Number of moles = mass / molar mass
Number of moles = 2.70 g / 182.81 g/mol = 0.0148 mol
Step 3: Since Ni(CN)2⋅4H2O produces 4 moles of water (H2O) per mole of Ni(CN)2⋅4H2O, we multiply the number of moles by 4.
Number of moles of H2O = 0.0148 mol x 4 = 0.0592 mol
Step 4: Convert the temperature from Celsius (°C) to Kelvin (K).
Temperature in Kelvin = 100.0 °C + 273.15 = 373.15 K
Now, we have all the information needed to calculate the volume.
PV = nRT
V = (nRT) / P
V = (0.0592 mol x 0.0821 L·atm/(mol·K) x 373.15 K) / 1.00 atm
V ≈ 1.79 L
Therefore, the volume of steam generated would be approximately 1.79 liters at 1.00 atm and 100.0 °C.
To find the volume of the steam, we need to use the ideal gas law equation, which is:
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, let's convert the given temperature from Celsius to Kelvin:
T = 100.0 °C + 273.15 = 373.15 K
Next, we need to determine the number of moles of water vapor that forms from the given sample of Ni(CN)2⋅4H2O.
1. Calculate the molar mass of Ni(CN)2⋅4H2O:
Ni: 1 atom x 58.69 g/mol = 58.69 g/mol
C: 2 atoms x 12.01 g/mol = 24.02 g/mol
N: 2 atoms x 14.01 g/mol = 28.02 g/mol
H: 8 atoms x 1.01 g/mol = 8.08 g/mol
O: 4 atoms x 16.00 g/mol = 64.00 g/mol
Molar mass of Ni(CN)2⋅4H2O = 58.69 g/mol + 24.02 g/mol + 28.02 g/mol + 8.08 g/mol + 64.00 g/mol = 182.81 g/mol
2. Calculate the number of moles:
moles = mass / molar mass
moles = 2.70 g / 182.81 g/mol = 0.0148 mol
Now, we can plug these values into the ideal gas law equation to find the volume (V):
PV = nRT
V = (nRT) / P
V = (0.0148 mol * 0.0821 L·atm/mol·K * 373.15 K) / 1.00 atm
Calculating the volume:
V = 4.83 L
So at 1.00 atm and 100.0 °C, the steam would occupy a volume of 4.83 liters.
mm = molar mass
grams H2O liberated = 2.70 g Ni(CN)2.4H2O x (4*mm H2O/mm Ni(CN)2.4H2O) = ? g H2O
Then mols H2O = grams H2O liberated/mm H2O and substitute into the following for n and solve for V in liters. Remember to convert T to kelvin which is 273 + 100 C = 373 K
Then use PV = nRT