What volumes will each of the following occupy at STP?
a. 1.80 x 10^24 molecules of SO3
b. 7.5 mol C2H6
c. 25.2 grams chlorine
is this right for a. 1.80 x 10^24 molecules of SO3(1 mol SO3/6.022E23)(22.4L/1 mol)? but then the book said the answer is 67.0 L SO3
You need to know the following for gases at STP. ONe mole will occupy 22.4 L and will contain 6.022E23 molecules. Those two facts will work almost any simple gas problem you have concerning volume and number of molecules.
To determine the volumes occupied at STP for each of the given substances, 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 (0.0821 L·atm/(K·mol)),
and T is the temperature in Kelvin.
At STP, the pressure is 1 atm and the temperature is 273.15 K.
a. 1.80 x 10^24 molecules of SO3:
First, we need to convert the number of molecules to moles using Avogadro's number, which states that 1 mole of any substance contains 6.022 x 10^23 particles.
So, the number of moles (n) of SO3 can be calculated as:
n = (1.80 x 10^24 molecules) / (6.022 x 10^23 molecules/mol)
n = 2.992 mol
Now, we can use the Ideal Gas Law to find the volume (V):
V = (nRT) / P
V = (2.992 mol)(0.0821 L·atm/(K·mol))(273.15 K) / (1 atm)
V ≈ 63.7 L
Therefore, 1.80 x 10^24 molecules of SO3 will occupy approximately 63.7 liters at STP.
b. 7.5 mol C2H6:
In this case, we already have the number of moles (n) of C2H6, so we can directly use the ideal gas law equation to find the volume (V):
V = (nRT) / P
V = (7.5 mol)(0.0821 L·atm/(K·mol))(273.15 K) / (1 atm)
V ≈ 168.3 L
Therefore, 7.5 moles of C2H6 will occupy approximately 168.3 liters at STP.
c. 25.2 grams chlorine:
To find the number of moles (n) of chlorine, we need to divide the given mass by the molar mass of chlorine, which is 35.45 g/mol.
n = (25.2 g) / (35.45 g/mol)
n ≈ 0.71 mol
Using the ideal gas law equation, we can find the volume (V):
V = (nRT) / P
V = (0.71 mol)(0.0821 L·atm/(K·mol))(273.15 K) / (1 atm)
V ≈ 16.1 L
Therefore, 25.2 grams of chlorine will occupy approximately 16.1 liters at STP.
To find the volumes that each of the given substances will occupy at STP (Standard Temperature and Pressure), we can use the ideal gas law and Avogadro's law.
The ideal gas law equation is:
PV = nRT
Where:
P = Pressure (at STP, it is 1 atm)
V = Volume
n = Number of moles
R = Ideal gas constant (0.0821 L·atm/(mol·K))
T = Temperature (at STP, it is 273 K)
Avogadro's law states that equal volumes of gases, at the same temperature and pressure, will contain an equal number of molecules (or moles).
Now let's proceed with the calculations for each part:
a. 1.80 x 10^24 molecules of SO3:
First, we need to convert the number of molecules to moles. Since we know that 1 mole of any substance contains 6.022 x 10^23 molecules (Avogadro's number), we can use this conversion factor:
1.80 x 10^24 molecules SO3 * (1 mole SO3 / 6.022 x 10^23 molecules SO3) = 2.99 moles SO3
Now, we can use the ideal gas law to find the volume:
V = (nRT) / P
V = (2.99 moles * 0.0821 L·atm/(mol·K) * 273 K) / 1 atm
V ≈ 67.8 L
Therefore, 1.80 x 10^24 molecules of SO3 will occupy approximately 67.8 liters at STP.
b. 7.5 mol C2H6:
Since we are already given the number of moles, we can directly use the ideal gas law to find the volume:
V = (nRT) / P
V = (7.5 moles * 0.0821 L·atm/(mol·K) * 273 K) / 1 atm
V ≈ 165.6 L
Therefore, 7.5 moles of C2H6 will occupy approximately 165.6 liters at STP.
c. 25.2 grams chlorine:
First, we need to convert the mass of chlorine to moles. We can use the molar mass of chlorine, which is 35.45 g/mol:
25.2 grams Cl2 * (1 mole Cl2 / 35.45 grams Cl2) ≈ 0.71 moles Cl2
Now, we can use the ideal gas law to find the volume:
V = (nRT) / P
V = (0.71 moles * 0.0821 L·atm/(mol·K) * 273 K) / 1 atm
V ≈ 16.5 L
Therefore, 25.2 grams of chlorine will occupy approximately 16.5 liters at STP.