a mixture of n2 and O2 contains 2.7 g of O2 and occupies 3.1 L at 28 degrees celcius, 810 mm Hg. how many grams on N2 are there
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To find the grams of N2 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.
First, let's convert the temperature from Celsius to Kelvin:
T = 28°C + 273.15 = 301.15 K
Next, let's convert the pressure from mmHg to atm:
P = 810 mmHg / 760 mmHg/atm = 1.0658 atm
We can now rearrange the ideal gas law equation to solve for the number of moles:
n = PV / RT
n = (1.0658 atm)(3.1 L) / (0.0821 L·atm/(mol·K))(301.15 K)
= 0.1341 mol
Now, we need to determine the moles of oxygen (O2) in the mixture. Since we have the mass of O2 and its molar mass, we can calculate the moles of O2.
The molar mass of O2 is 32 g/mol.
n(O2) = mass(O2) / molar mass(O2)
= 2.7 g / 32 g/mol
= 0.0844 mol
Since the mixture contains both N2 and O2, the total moles of gas in the mixture can be obtained by summing up the moles of N2 and O2:
Total moles = n(N2) + n(O2)
We want to find the grams of N2, so let's rearrange the equation to solve for n(N2):
n(N2) = Total moles - n(O2)
= 0.1341 mol - 0.0844 mol
= 0.0497 mol
Finally, we can convert the moles of N2 to grams using its molar mass. The molar mass of N2 is 28 g/mol.
grams(N2) = n(N2) × molar mass(N2)
= 0.0497 mol × 28 g/mol
= 1.39 g
Therefore, there are approximately 1.39 grams of N2 in the mixture.