1. What volume will a mixture of 5.00 mol of H2 and 0.500 mol of CO2 occupy at STP?
Use PV = nRT. Substitute 5.5 for n.
Here's my work please tell me if its right DrBob222;
P=760 torr or 1 atm
V=?
T= 273 K
n= 5.00 + 0.500= 5.5 mole
V= nRT/P
= (5.5 mol)(0.0821 L x atm/mol*K)(273 K)
divided by 1 atm
= 123.27 L
To determine the volume of a gas mixture at standard temperature and pressure (STP), you 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
R is the ideal gas constant (0.0821 L.atm/mol.K)
T is the temperature in Kelvin (273.15 K at STP)
First, calculate the total number of moles in the mixture by adding the number of moles of each gas:
Total moles = 5.00 mol (H2) + 0.500 mol (CO2) = 5.50 mol
Now, rearrange the ideal gas law equation to solve for volume:
V = (nRT) / P
Substituting the known values:
V = (5.50 mol) * (0.0821 L.atm/mol.K) * (273.15 K) / (1 atm)
V ≈ 123.1 L
Therefore, the mixture of 5.00 mol of H2 and 0.500 mol of CO2 will occupy approximately 123.1 liters at STP.
To find the volume of the mixture of gases at STP, you can use the ideal gas law equation. The ideal gas law is given by:
PV = nRT,
where:
P is the pressure (in this case, at STP it is 1 atm),
V is the volume,
n is the number of moles of the gas,
R is the ideal gas constant (0.0821 L·atm/mol·K),
T is the temperature (at STP, it is 273.15 K).
In this case, you have a mixture of H2 and CO2, so you need to calculate the volumes separately and then add them together.
For H2:
n = 5.00 mol
P = 1 atm
T = 273.15 K
R = 0.0821 L·atm/mol·K
Plugging in these values into the ideal gas law equation for H2:
V1 = (n1 * R * T) / P,
V1 = (5.00 mol * 0.0821 L·atm/mol·K * 273.15 K) / 1 atm,
V1 = 113.08 L.
For CO2:
n = 0.500 mol
P = 1 atm
T = 273.15 K
R = 0.0821 L·atm/mol·K
Plugging in these values into the ideal gas law equation for CO2:
V2 = (n2 * R * T) / P,
V2 = (0.500 mol * 0.0821 L·atm/mol·K * 273.15 K) / 1 atm,
V2 = 11.31 L.
Now to find the total volume of the mixture, you add the volumes of H2 and CO2:
V_total = V1 + V2,
V_total = 113.08 L + 11.31 L
V_total = 124.39 L.
Therefore, the volume of the mixture of 5.00 mol of H2 and 0.500 mol of CO2 at STP is 124.39 L.