What volume will the following quantities of gases occupy at STP?
5.0 g of H(2) subscript=2
228.0 g of N(2) subscript=2
Convert the grams to moles by dividing the grams by the formula mass.
Multiply the number of moles by 22.4 L/mol.
To determine the volume that each gas will occupy at STP (Standard Temperature and Pressure), we can use the ideal gas law equation:
PV = nRT
Where:
P is the pressure (at STP, it is 1 atm)
V is the volume
n is the number of moles of gas
R is the ideal gas constant (0.0821 L·atm/(mol·K))
T is the temperature (at STP, it is 273.15 K)
First, we need to calculate the number of moles for each gas.
For hydrogen gas (H(2)):
1 mole of H(2) has a molar mass of approximately 2 g (1 gram per mole).
So, to calculate the number of moles, we divide the mass by the molar mass:
5.0 g H(2) / 2 g/mol = 2.5 mol H(2)
For nitrogen gas (N(2)):
1 mole of N(2) has a molar mass of approximately 28 g (14 g per mole).
So, to calculate the number of moles, we divide the mass by the molar mass:
228.0 g N(2) / 28 g/mol = 8.14 mol N(2)
Now that we have the number of moles, we can use the ideal gas law equation to calculate the volumes.
For hydrogen gas (H(2)):
PV = nRT
Since the pressure (P) is 1 atm, the temperature (T) is 273.15 K, and the number of moles (n) is 2.5 mol, we can rearrange the ideal gas law equation:
V = nRT / P
V = (2.5 mol)(0.0821 L·atm/(mol·K))(273.15 K) / 1 atm
V ≈ 56.16 L
Therefore, 5.0 g of H(2) at STP will occupy approximately 56.16 liters of volume.
For nitrogen gas (N(2)):
Using the same approach:
V = nRT / P
V = (8.14 mol)(0.0821 L·atm/(mol·K))(273.15 K) / 1 atm
V ≈ 187.9 L
Therefore, 228.0 g of N(2) at STP will occupy approximately 187.9 liters of volume.