The total volume of mixture of 2grams of helium and 7 grams of nitrogen at s.t.p is 1)22.4 2)11.2 3)16.8 4)5.6 in liters help me please.
To find the total volume of the mixture of helium and nitrogen, we first need to calculate the number of moles of each gas.
1. Calculate the moles of helium:
We can use the ideal gas 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.
At standard temperature and pressure (STP), the pressure is 1 atmosphere (atm), the temperature is 273 Kelvin (K), and the ideal gas constant is 0.0821 L·atm/(mol·K).
Using the equation n = m/M, where n is the number of moles, m is the mass, and M is the molar mass, we can determine the moles of helium.
Given that the mass of helium is 2 grams and the molar mass of helium is approximately 4 grams/mol, we have:
n(He) = 2 g / 4 g/mol = 0.5 mol
2. Calculate the moles of nitrogen:
Using the same equation n = m/M, we can determine the moles of nitrogen.
Given that the mass of nitrogen is 7 grams and the molar mass of nitrogen is approximately 28 grams/mol, we have:
n(N2) = 7 g / 28 g/mol = 0.25 mol
3. Add the moles of each gas:
n(total) = n(He) + n(N2)
n(total) = 0.5 mol + 0.25 mol = 0.75 mol
4. Calculate the total volume:
We can again use the ideal gas equation, PV = nRT, to determine the total volume.
Given that the pressure is 1 atm, the number of moles is 0.75 mol, the temperature is 273 K, and the ideal gas constant is 0.0821 L·atm/(mol·K), we have:
V = (nRT) / P
V = (0.75 mol * 0.0821 L·atm/(mol·K) * 273 K) / 1 atm
V ≈ 16.8 liters
Therefore, the total volume of the mixture of helium and nitrogen at STP is approximately 16.8 liters.
The correct answer is option 3) 16.8.
To find the total volume of the mixture, we need to calculate the individual volumes of helium and nitrogen separately, and then add them together.
First, let's calculate the volume of helium. 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.
At standard temperature and pressure (STP), the pressure is 1 atmosphere (atm) and the temperature is 0 degrees Celsius or 273.15 Kelvin (K).
To calculate the number of moles of helium, we use the equation: moles = mass / molar mass. The molar mass of helium is 4 grams/mol.
moles of helium = 2 grams / 4 grams/mol = 0.5 moles
Now we can calculate the volume of helium using the ideal gas law:
V = (n * R * T) / P
= (0.5 moles * 0.0821 L*atm/(mol*K) * 273.15 K) / 1 atm
= 11.2 liters
So, the volume of helium is 11.2 liters.
Next, let's calculate the volume of nitrogen. Follow the same process as above.
moles of nitrogen = 7 grams / 28 grams/mol (molar mass of nitrogen)
= 0.25 moles
V = (n * R * T) / P
= (0.25 moles * 0.0821 L*atm/(mol*K) * 273.15 K) / 1 atm
= 5.6 liters
So, the volume of nitrogen is 5.6 liters.
Finally, to find the total volume of the mixture, we add the volumes of helium and nitrogen:
Total volume = volume of helium + volume of nitrogen
= 11.2 liters + 5.6 liters
= 16.8 liters
Therefore, the total volume of the mixture of 2 grams of helium and 7 grams of nitrogen at STP is 16.8 liters.
you have 1/2 mole of He, 1/2 mole of N2, total one mole of gas mixture. At STP, that is 22.4 dm^3. I recommend you memerorize that.
Now if you want to work it out
PV=nRT
V=nRT/P put in STP numbers, N at one mole, and you will get a volume of 0.0223m^3 or 22.4 dm^3 or what folks living in the past century called 22.4 liters. If you wish, your teacher could teach it to you in gallons, or Imperial gallons, or cubic centimeters, or even bushels. 22.4 dm^3