use the molar volume of a gas to solve the following at STP:
a. the number of moles of O2 in 44.8 L O2 gas.
b. the number of moles of CO2 in 4.00L CO2 gas.
c. the volume(L) of 6.40g O2.
d. the volume(mL) occupied by 50.0g neon.
Just remember the molar volume of a gas at STP is 22.4L; i.e., 1 mole of a gas @ STP will occupy 22.4 L.
Where grams is given, moles = grams/molar mass.
To solve these problems using the molar volume of a gas at STP, we need to know the molar volume value. At STP (Standard Temperature and Pressure), the molar volume of any ideal gas is approximately 22.4 liters per mole (L/mol).
a. To find the number of moles of O2 in 44.8 L of O2 gas:
Given: Volume of O2 gas = 44.8 L
Using the molar volume concept, we can set up the following equation:
44.8 L (1 mol/22.4 L) = 2.00 mol
Therefore, there are 2.00 moles of O2 in 44.8 L of O2 gas.
b. To find the number of moles of CO2 in 4.00 L of CO2 gas:
Given: Volume of CO2 gas = 4.00 L
Using the molar volume concept again, we can set up the equation:
4.00 L (1 mol/22.4 L) = 0.1786 mol
Therefore, there are approximately 0.1786 moles of CO2 in 4.00 L of CO2 gas.
c. To find the volume (in liters) of 6.40 g of O2:
Given: Mass of O2 gas = 6.40 g
First, we need to convert grams to moles using the molar mass of O2.
The molar mass of O2 = 32.00 g/mol (16.00 g/mol for each oxygen atom).
Number of moles = Mass (g) / Molar Mass (g/mol)
Number of moles = 6.40 g / 32.00 g/mol = 0.2000 mol
Now, we can use the molar volume concept as follows:
Volume = Number of moles × Molar volume
Volume = 0.2000 mol × 22.4 L/mol = 4.48 L
Therefore, 6.40 g of O2 gas occupies a volume of 4.48 L at STP.
d. To find the volume (in mL) occupied by 50.0 g of neon:
Given: Mass of neon gas = 50.0 g
First, convert grams to moles using the molar mass of neon.
The molar mass of neon = 20.18 g/mol
Number of moles = Mass (g) / Molar Mass (g/mol)
Number of moles = 50.0 g / 20.18 g/mol = 2.4742 mol
Now, use the molar volume concept:
Volume = Number of moles × Molar volume
Volume = 2.4742 mol × 22.4 L/mol = 55.40 L
To convert liters to milliliters, multiply by 1000 mL/L:
Volume = 55.40 L × 1000 mL/L = 55400 mL
Therefore, 50.0 g of neon gas occupies a volume of 55400 mL at STP.
To solve the given problems using the molar volume of a gas at STP (Standard Temperature and Pressure), we need to use the concept of the ideal gas law and Avogadro's principle.
The molar volume of a gas at STP is defined as 22.4 liters per mole. This means that at STP, one mole of any gas occupies a volume of 22.4 liters.
Let's solve each problem step by step.
a. To find the number of moles of O2 in 44.8 L of O2 gas:
Using the molar volume of a gas at STP (22.4 L/mol), we can set up a proportion to solve for moles of O2:
44.8 L O2 gas = X moles of O2
X = (44.8 L O2 gas) / (22.4 L/mol)
X = 2 moles
Therefore, there are 2 moles of O2 in 44.8 L of O2 gas at STP.
b. To find the number of moles of CO2 in 4.00 L of CO2 gas:
Using the same approach as above:
4.00 L CO2 gas = X moles of CO2
X = (4.00 L CO2 gas) / (22.4 L/mol)
X = 0.179 moles
Therefore, there are 0.179 moles of CO2 in 4.00 L of CO2 gas at STP.
c. To find the volume (in liters) of 6.40 g of O2:
First, we need to convert grams to moles using the molar mass of O2.
The molar mass of oxygen (O2) is approximately 32 g/mol.
Number of moles = Mass / Molar mass
Number of moles = 6.40 g / 32 g/mol
Number of moles = 0.2 moles
Since 1 mole of gas occupies 22.4 L of volume at STP, we can find the volume of 0.2 moles of O2:
Volume = (0.2 moles) * (22.4 L/mol)
Volume = 4.48 L
Therefore, 6.40 g of O2 occupies 4.48 L of volume at STP.
d. To find the volume (in milliliters) occupied by 50.0 g of neon:
First, we need to convert grams to moles using the molar mass of neon.
The molar mass of neon (Ne) is approximately 20 g/mol.
Number of moles = Mass / Molar mass
Number of moles = 50.0 g / 20 g/mol
Number of moles = 2.5 moles
Since 1 mole of gas occupies 22.4 L of volume at STP, we can find the volume of 2.5 moles of neon:
Volume = (2.5 moles) * (22.4 L/mol)
Volume = 56 L
Since 1 L = 1000 mL, we convert the volume to milliliters:
Volume = 56 L * 1000 mL/L
Volume = 56,000 mL
Therefore, 50.0 g of neon occupies 56,000 mL of volume at STP.