A 3.31-g sample of lead nitrate, , molar mass = 331 g/mol, is heated in an
evacuated cylinder with a volume of 2.37 L. The salt decomposes when heated, according to
the equation:
Assuming complete decomposition, what is the pressure in the cylinder after decomposition
and cooling to a temperature of 300. K? Assume the takes up negligible volume.
You didn't write the decomposition equation nor did you not if was Pb(II) nitrate or Pb(IV) nitratem. Is this it?
2Pb(NO3)2→2PbO+4NO2+O2
mols Pb(NO3)2 = grams/molar mass = 3.31/331 = 0.01
mols NO2 formed = 4 x 0.01 = 0.04
mols O2 formed = 1 x 0.01 = 0.01
Total mols gas = 0.05
Then use PV = nRT and solve for total pressure in the container. Post your work if you get stuck.
Well, let's break it down, because that's what things tend to do when they decompose. First, we need to find the number of moles of lead nitrate that we have.
We can do this by taking the mass of the sample (3.31 g) and dividing it by the molar mass of lead nitrate (331 g/mol).
So, 3.31 g / 331 g/mol = 0.01 mol.
Now, according to Avogadro's Law, equal volumes of gases at the same temperature and pressure contain the same number of molecules.
Since the volume of the cylinder is given as 2.37 L, the number of moles of gas after decomposition will still be 0.01 mol.
Now, we can use the ideal gas law, which states that PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant (0.08206 L·atm/(mol·K)), and T is the temperature in Kelvin.
Plugging in the values, we get P * 2.37 = 0.01 * 0.08206 * 300.
After some calculations, we find P = 0.104 atm.
So, the pressure in the cylinder after decomposition and cooling to a temperature of 300 K is approximately 0.104 atmospheres. Now, that's a pretty low-pressure situation. It's almost as if the molecules are whispering to each other instead of bumping into each other. Maybe they're talking about their hopes and dreams, who knows? The point is, it's a pretty relaxed environment in that cylinder.
To determine the pressure in the cylinder after decomposition and cooling, 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 gas constant, and T is the temperature.
First, let's calculate the number of moles of lead nitrate in the sample.
Given:
Mass of lead nitrate = 3.31 g
Molar mass of lead nitrate = 331 g/mol
Number of moles = mass / molar mass
Number of moles = 3.31 g / 331 g/mol
Number of moles = 0.01 mol
Next, let's calculate the pressure.
Given:
Volume = 2.37 L
Temperature = 300 K
Rearranging the ideal gas law equation, we get:
P = (nRT) / V
P = (0.01 mol * 0.0821 L·atm/mol·K * 300 K) / 2.37 L
P = (0.2463 L·atm) / 2.37 L
P ≈ 0.104 atm
Therefore, the pressure in the cylinder after decomposition and cooling to a temperature of 300 K is approximately 0.104 atm.
To find the pressure in the cylinder after decomposition and cooling, we need to use the ideal gas law equation, which states:
PV = nRT
Where:
P = pressure (in units of pressure, such as atm or Pa)
V = volume (in units of volume, such as L or m^3)
n = number of moles of gas
R = ideal gas constant (0.0821 L·atm/mol·K)
T = temperature (in units of temperature, such as K or °C)
First, we need to calculate the number of moles of gas produced by the decomposition of lead nitrate. To do this, we can use the molar mass and the mass of the lead nitrate given.
Given:
Mass of lead nitrate (m) = 3.31 g
Molar mass of lead nitrate (M) = 331 g/mol
Number of moles (n) = mass / molar mass
n = 3.31 g / 331 g/mol
Now, we can substitute the values into the ideal gas law equation:
PV = nRT
P * 2.37 L = (3.31 g / 331 g/mol) * (0.0821 L·atm/mol·K) * 300 K
Simplifying the equation, we can calculate the value of P:
P = (3.31 g / 331 g/mol) * (0.0821 L·atm/mol·K) * 300 K / 2.37 L
Calculating the above expression will give you the value of the pressure (P) in the cylinder after decomposition and cooling to a temperature of 300 K.