At STP, a container holds 1.01 g of hydrogen and 64.00 g of oxygen gas. What's the container volume?
In the container we have 1.01 g of H₂ and 64.00 g of O₂. The first step to solve the problem is to find the number of moles of each gas that we have in the container. To do that we have to find the molar mass of each gas. We have to look for the atomic mass of each element in the periodic table.
atomic mass of H: 1.01 amu atomic mass of O: 16.00 amu
Using those values we can find the molar mass of H₂ and the molar mass of O₂.
molar mass of H₂ = 2 * 1.01
molar mass of H₂ = 2.02 g/mol
molar mass of O₂ = 2 * 16.00
molar mass of O₂ = 32.00 g/mol
Using the molar mass of each gas we can find the number of moles of each gas that we have in the container.
mass of H₂ = 1.01 g mass of O₂ = 64.00 g
number of moles of H₂ = mass of H₂/(molar mass of H₂)
number of moles of H₂ = 1.01 g/(2.02 g/mol) = 1.00 mol
number of moles of H₂ = 0.50 mol
number of moles of O₂ = mass of O₂/(molar mass of O₂)
number of moles of O₂ = 64.00 g/(32.00 g/mol)
number of moles of O₂ = 2.00 moles
So we found that the container holds 2.00 moles of oxygen and 0.50 mol of hydrogen at STP. Using the ideal gas law we can find the volume of the container.
n = total number of moles = number of moles of O₂ + number of moles of H₂
n = 2.00 moles + 0.50 moles
n = 2.50 moles
The container is at STP, that means: P = 1.00 atm T = 273.15 K
P * V = n * R * T
V = n * R * T / P
V = 2.50 moles * (0.082 atm*L/mol*K) * 273.15 K / ( 1.00 atm)
V = 56.0 L
P * V = n * R * T
V = n * R * T / P
V = 2.50 moles * (8.314 kPa*L/mol*K) * 273.15 K / ( 101.325 kPa)
V= 56.0 L
Answer: The volume of the container is 56.0 L