A special gas mixture is used in bacterial growth chambers which contains 5.00% by mass of CO2 and 95.0% O2. If the mole fraction of CO2 and O2 in the mixture is 0.0998 and 0.2340, respectively. What is the partial pressure in atmospheres of O2 and CO2 gases at a total pressure of 0.6880 atm?
First, we can use the mole fractions to calculate the number of moles of each gas present in the mixture. Let's assume we have 100 grams of the mixture, then:
- Moles of CO2 = 0.05 x 100 / 44.01 = 0.1136 moles
- Moles of O2 = 0.95 x 100 / 32.00 = 2.9688 moles
Next, we can use the total pressure and mole fractions to calculate the partial pressures of each gas. The partial pressure of a gas is equal to its mole fraction times the total pressure. So:
- Partial pressure of CO2 = 0.0998 x 0.6880 atm = 0.0685 atm
- Partial pressure of O2 = 0.2340 x 0.6880 atm = 0.1608 atm
Therefore, the partial pressure of CO2 in the mixture is 0.0685 atm and the partial pressure of O2 is 0.1608 atm.
If the theoretical yield of PbCl2 is 16.547g from the reaction of 7.000g of NaCl with 50.400g of Pb(NO3)2. Calculate the % yield of PbCl2 if 12.788g of PbCl2 was actually obtained? (Atomic mass: Na-22.990 g/mol; Cl-35.453g/mol; 0-15.999g/mol; Pb-207.200 g/mol; N-14.007).
First, let's balance the equation:
2NaCl + Pb(NO3)2 → 2NaNO3 + PbCl2
From the balanced equation, the stoichiometry of the reaction is 2 moles of NaCl to 1 mole of Pb(NO3)2 to 1 mole of PbCl2.
We can calculate the number of moles of Pb(NO3)2 used in the reaction as follows:
50.400 g Pb(NO3)2 × 1 mol Pb(NO3)2 / 331.209 g Pb(NO3)2 = 0.1521 mol Pb(NO3)2
Therefore, the number of moles of PbCl2 that would be produced if the reaction went to completion is also 0.1521 mol (1:1 stoichiometry with Pb(NO3)2). The theoretical yield of PbCl2 can be calculated from the molar mass:
16.547 g PbCl2 × 1 mol PbCl2 / 278.104 g PbCl2 = 0.0595 mol PbCl2
Now we can calculate the percent yield:
Percent yield = actual yield / theoretical yield × 100%
= 12.788 g / 16.547 g × 100%
= 77.29%
Therefore, the percent yield of PbCl2 is 77.29%.
To find the partial pressure of CO2 and O2 gases, we need to calculate the pressure exerted by each gas individually.
Step 1: Calculate the mole fraction of CO2 and O2.
Mole fraction (X) is calculated by dividing the number of moles of a component by the total number of moles in the mixture. We are given the mole fraction for CO2 and O2.
Mole fraction of CO2 (X_CO2) = 0.0998
Mole fraction of O2 (X_O2) = 0.2340
Step 2: Calculate the number of moles of CO2 and O2.
We know that the sum of mole fractions equals 1.
X_CO2 + X_O2 = 1
Since X_CO2 is given as 0.0998, we can substitute this value and solve for X_O2.
0.0998 + X_O2 = 1
X_O2 = 1 - 0.0998
X_O2 = 0.9002
Now we can calculate the number of moles of each component using the mole fraction and the total pressure of the mixture.
Step 3: Calculate the number of moles.
PV = nRT
For CO2:
Partial pressure of CO2 (P_CO2) = X_CO2 * Total pressure
= 0.0998 * 0.6880 atm
For O2:
Partial pressure of O2 (P_O2) = X_O2 * Total pressure
= 0.9002 * 0.6880 atm
Step 4: Convert moles to pressure in atmospheres.
The ideal gas law gives us the equation PV = nRT.
We can rearrange this equation to solve for pressure.
Pressure (P) = n * R * T / V
Since we have the number of moles of each component, we can plug this into the equation to calculate the partial pressure of each gas.
Using the ideal gas constant R = 0.0821 L * atm / mol * K, and assuming temperature (T) and volume (V) are constant for both gases, we can calculate the partial pressures.
Step 5: Calculate pressure exerted by CO2 and O2.
Pressure (P) = n * R * T / V
Let's assume T and V are constants for both gases.
For CO2:
P_CO2 = (n_CO2 * R * T) / V
For O2:
P_O2 = (n_O2 * R * T) / V
Since the temperature and volume are the same for both gases, we can ignore those variables.
So, the partial pressure of CO2 is given by P_CO2 = (n_CO2 * R) / V
And the partial pressure of O2 is given by P_O2 = (n_O2 * R) / V
Now we can proceed with the calculation.
Step 6: Calculate the partial pressure of CO2 and O2.
P_CO2 = (n_CO2 * R) / V
= (0.0998 * 0.6880 atm) / (0.0821 L * atm / mol * K)
P_O2 = (n_O2 * R) / V
= (0.9002 * 0.6880 atm) / (0.0821 L * atm / mol * K)
To calculate the partial pressure of a gas in a mixture, you can use the equation:
Partial Pressure = Mole Fraction × Total Pressure
Given that the total pressure is 0.6880 atm, we need to calculate the partial pressure of CO2 and O2 gases.
1. First, let's calculate the mole fractions of CO2 and O2:
Mole fraction of CO2 = 0.0998
Mole fraction of O2 = 0.2340
2. Now, we can calculate the partial pressures of CO2 and O2:
Partial Pressure of CO2 = Mole Fraction of CO2 × Total Pressure
= 0.0998 × 0.6880 atm
= 0.06862 atm
Partial Pressure of O2 = Mole Fraction of O2 × Total Pressure
= 0.2340 × 0.6880 atm
= 0.161232 atm
Therefore, the partial pressure of CO2 in the gas mixture is approximately 0.06862 atm, and the partial pressure of O2 is approximately 0.161232 atm.