15.8 grams of potassium chlorate is decompossed and the gas is colected over water at 29 degrees celcius when water vapor pressure is 30 mmHg. Assuming 100% yeild of products, what is the total pressure in the vessel after the reaction in complete?
2KClO3 ==> 2KCl + 3O2
mols KClO3 = 15.8g/molar mass
Using the coefficients in the balanced equation, convert mols KClO3 to mols O2. That is mols KClO3 x (3 mols O2/2 mols KClO3) = ?
Then PV = nRT to convert mols KClO3 to pressure in atm. I would convert that to mm Hg(multiply n*760) and add the vapor pressure of the H2O for total P.
what do you use for the V in PV=nRT?
I don't know and I didn't noticed when I posted the response that it was not listed. You must have it or have some way to calculate it. Do you have any other information about the problem.
no
The problem is impossible to do. There must have been a grammatical error somewhere
To find the total pressure in the vessel after the reaction is complete, we need to consider the pressure contribution of both the collected gas and the water vapor.
First, let's calculate the pressure contribution of the collected gas. Since the gas is collected over water, we need to take into account the vapor pressure of water at 29 degrees Celsius, which is 30 mmHg. This means that the pressure displayed by the gas collection vessel will contain the partial pressure of the gas and the partial pressure of the water vapor.
Now, according to Dalton's Law of Partial Pressures, the total pressure is equal to the sum of the partial pressures of each gas. Therefore, we can calculate the partial pressure of the collected gas by subtracting the vapor pressure of water from the total pressure.
To find the number of moles of the collected gas, we need to use the molar mass of potassium chlorate (KClO3), which is 122.55 g/mol. Dividing the given mass of potassium chlorate (15.8 g) by its molar mass will give us the number of moles.
15.8 g KClO3 ÷ 122.55 g/mol = 0.129 mol KClO3
Since potassium chlorate decomposes according to the following equation:
2 KClO3(s) → 2 KCl(s) + 3 O2(g)
We can see that for every 2 moles of KClO3 decomposed, 3 moles of O2 are produced. Therefore, for 0.129 mol of KClO3 decomposed, we will produce (0.129 mol O2) x (3 mol O2/2 mol KClO3) = 0.1945 mol O2.
Now we can calculate the partial pressure of the collected gas (O2) using the ideal gas law. The ideal gas law equation is:
PV = nRT
Where:
P = pressure (in atm)
V = volume (in liters)
n = number of moles
R = ideal gas constant (0.0821 L.atm/mol.K)
T = temperature in Kelvin (29 °C + 273.15 = 302.15 K)
We have the number of moles (0.1945 mol O2), and since the gas is collected over water, we can assume the volume of the gas is equal to the volume of the gas collection vessel.
Now, let's find the partial pressure of the collected gas at 29 degrees Celsius:
P(O2) = (n(O2) x R x T) / V
Substitute the values into the equation:
P(O2) = (0.1945 mol x 0.0821 L.atm/mol.K x 302.15 K) / V
Now, let's calculate the total pressure in the vessel:
Total pressure = Partial pressure of the collected gas (O2) + Vapor pressure of water
Total pressure = P(O2) + vapor pressure of water = P(O2) + 30 mmHg
Please note that to find the exact total pressure, the volume of the gas collection vessel (V) is required. Once you have the value for V, substitute it into the equation to obtain the total pressure in the vessel after the reaction is complete.