A gas sample at STP contains 1.15 oxygen and 1.55g nitrogen. What is the volume of the gas sample?
the answer 2.04. I want to know how to get the that answer
figure the moles.
1 mole occupies 22.4L
1.15g O2 is 0.0359 moles O2
1.55g N2 is 0.0554 moles N2
A total of 0.091 moles * 22.4L/mole = 2.04 L
The 22.4L is from STP if your curious
Well, let's get to the bottom of this gas situation together! To find the volume of the gas sample, we first need to figure out the number of moles of each gas present.
For oxygen (O2), we can use the molar mass of oxygen (32 g/mol) to convert the given mass (1.15 g) into moles:
1.15 g O2 * (1 mol O2 / 32 g O2) = 0.036 mol O2
For nitrogen (N2), we'll go through the same process:
1.55 g N2 * (1 mol N2 / 28 g N2) = 0.055 mol N2
Now that we know the number of moles of each gas in the sample, we can use the ideal gas law to find the volume. At STP (standard temperature and pressure), 1 mole of gas occupies 22.4 liters. So, we'll multiply the number of moles of the total gas by 22.4 L/mol to get the volume:
(0.036 mol O2 + 0.055 mol N2) * 22.4 L/mol = 2.04 L
And there you have it! The volume of the gas sample is approximately 2.04 liters. I hope this helps!
To find the volume of the gas sample, we need to use the ideal gas law equation, which is PV = nRT. In this equation, P represents the pressure, V represents the volume, n represents the number of moles, R represents the ideal gas constant, and T represents the temperature.
At STP (Standard Temperature and Pressure), the pressure is 1 atm, the temperature is 273 K (0 degrees Celsius), and the ideal gas constant is 0.0821 L·atm/K·mol.
First, we need to find the number of moles of oxygen and nitrogen in the gas sample:
Moles of oxygen = mass / molar mass = 1.15 g / 32.00 g/mol (molar mass of oxygen)
Moles of oxygen = 0.036 mol
Moles of nitrogen = mass / molar mass = 1.55 g / 28.01 g/mol (molar mass of nitrogen)
Moles of nitrogen = 0.055 mol
Next, we add the moles of the two gases together to get the total moles:
Total moles = moles of oxygen + moles of nitrogen
Total moles = 0.036 mol + 0.055 mol
Total moles = 0.091 mol
Now, we can substitute the values into the ideal gas law equation to find the volume:
V = (nRT) / P
V = (0.091 mol)(0.0821 L·atm/K·mol)(273 K) / 1 atm
V = 2.041 L
Therefore, the volume of the gas sample is approximately 2.04 L.
To find the volume of the gas sample, we can use the ideal gas law equation: PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.
Given that the gas sample is at STP (Standard Temperature and Pressure), we know that the pressure is 1 atm and the temperature is 273.15 K. We also have the masses of oxygen and nitrogen in the sample.
The first step is to convert the masses of oxygen and nitrogen to moles.
To convert from grams to moles, we use the molar mass of each element.
The molar mass of oxygen (O2) is approximately 32 g/mol, and the molar mass of nitrogen (N2) is approximately 28 g/mol.
For oxygen:
moles of oxygen = mass of oxygen / molar mass of oxygen
moles of oxygen = 1.15 g / 32 g/mol
For nitrogen:
moles of nitrogen = mass of nitrogen / molar mass of nitrogen
moles of nitrogen = 1.55 g / 28 g/mol
Next, we add the moles of oxygen and nitrogen to find the total number of moles in the gas sample.
Total moles = moles of oxygen + moles of nitrogen
Now that we have the number of moles, we can rearrange the ideal gas law equation to solve for volume:
V = (nRT) / P
Substituting the known values:
V = (Total moles * R * T) / P
Now we can calculate the volume by plugging in the values:
V = (Total moles * 0.0821 L/mol*K * 273.15 K) / 1 atm
Once you perform the calculations, you should obtain the final answer of approximately 2.04 Liters (L) as the volume of the gas sample.