A mixture of n2 and o2 contains 2.7 grams of O2 and occupies 3.1 L at 28 degrees Celsius, 810 mmHg. how many grams of N2 are present?
Use PV = nRT to calculate partial pressure of O2.
partial pressure O2 + partial pressure N2 = Ptotal = 810 mm Hg. Solve for partial pressure N2.
Use PV = nRT to solve for n = number of moles N2, then convert to grams.
what are the partial pressures of N2 and O2 in air at .8atm? this is the only info avaiable
To find the number of grams of N2 present in the mixture, we can use the ideal gas law equation: PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.
First, let's convert the temperature from Celsius to Kelvin:
T(K) = 28°C + 273.15 = 301.15 K
Next, we can calculate the number of moles of O2 present using the given information. We know the volume, pressure, and temperature, so we can rearrange the ideal gas law equation to solve for the number of moles, n:
n = PV / RT
Let's substitute the values into the equation:
n(O2) = (810 mmHg)(3.1 L) / (0.0821 L·atm/(mol·K))(301.15 K)
Note: We need to convert the pressure from mmHg to atm for consistency. 1 atm is equal to 760 mmHg.
n(O2) = (810 mmHg)(3.1 L) / (0.0821 L·atm/(mol·K))(301.15 K) × (1 atm / 760 mmHg)
n(O2) ≈ 0.1192 moles
Now, since the ratio of N2 to O2 in the mixture is 1:1, the number of moles of N2 present is also approximately 0.1192 moles.
Finally, to calculate the grams of N2, we can use the molar mass of N2, which is approximately 28.01 g/mol.
m(N2) = n(N2) × molar mass(N2)
m(N2) ≈ 0.1192 moles × 28.01 g/mol
Therefore, the approximate amount of N2 present in the mixture is 3.33 grams.