Solid copper oxide reacts with ammonia gas according to the following equation: 3CuO + 2NH3 -> 3Cu + N2 + 3H20
50 grams of CuO is placed in an 80L reaction vessel, and the vessel was evacuated. Ammonia gas was then gradually introduced into the vessel, slow enough for the reaction to proceed, until the total pressure in the vessel (after the reaction) was 1 atm. What are the partial pressures of all three gases in the vessel? (Temperature is 180 degrees celsius).
I was thinking of getting each of the moles, and then using PV-nRT for each one. Is that correct? Or is there a shorter way to do it? I was also thinking instead of using the ratio nA/ntotal= PA/Ptotal but I'm not sure how to go about it.
Thanks!
I would do this.
Use PV = nRT. You know P, V, R, and T. Solve for n = total mols.
Then use stoichiometry to calculate mols CuO that reacts and from that you get mols H2 and mols N2. Since you know
mols H2, mols N2, and total mols, you can subtract and find mols NH3.
Then use PV = nRT for each n of each gas and solve for p which will be the partial pressure of the gas.
To find the partial pressures of the gases in the reaction vessel, you 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 in Kelvin.
First, let's calculate the number of moles of CuO (copper oxide) present. The molar mass of CuO is 63.55 + 16 = 79.55 g/mol, so 50 grams of CuO is equal to:
moles of CuO = mass / molar mass = 50 g / 79.55 g/mol = 0.628 mol
Next, let's use the stoichiometry of the reaction to determine the number of moles of each gas produced.
According to the balanced equation:
3 CuO + 2 NH3 -> 3 Cu + N2 + 3 H2O
The stoichiometric ratio between CuO and NH3 is 3:2. Therefore, the number of moles of NH3 that reacts will be:
moles of NH3 = (3/2) * moles of CuO = (3/2) * 0.628 mol = 0.942 mol
Now, we calculate the total number of moles of gas in the reaction vessel after the reaction:
Total moles of gas = moles of Cu (produced) + moles of N2 (produced) + moles of H2O (produced) + moles of NH3 (reacted)
From the balanced equation, we can see that 3 moles of Cu, 1 mole of N2, and 3 moles of H2O are produced for every 2 moles of NH3 reacted. Therefore,
Total moles of gas = (3/2) * moles of NH3 (reacted) = (3/2) * 0.942 mol = 1.413 mol
Now, let's compute the partial pressure of each gas using the ideal gas law.
The temperature given is 180 degrees Celsius, which is 180 + 273.15 = 453.15 Kelvin.
For each gas, we can use PV = nRT to find the partial pressure:
Partial pressure of Cu = (moles of Cu / Total moles of gas) * Total pressure
Partial pressure of Cu = (3/1.413) * 1 atm = 2.122 atm
Partial pressure of N2 = (moles of N2 / Total moles of gas) * Total pressure
Partial pressure of N2 = (1/1.413) * 1 atm = 0.708 atm
Partial pressure of H2O = (moles of H2O / Total moles of gas) * Total pressure
Partial pressure of H2O = (3/1.413) * 1 atm = 2.122 atm
Partial pressure of NH3 = (moles of NH3 / Total moles of gas) * Total pressure
Partial pressure of NH3 = (0.942/1.413) * 1 atm = 0.666 atm
So, the partial pressures of the three gases in the vessel after the reaction are:
Partial pressure of Cu: 2.122 atm
Partial pressure of N2: 0.708 atm
Partial pressure of H2O: 2.122 atm
Partial pressure of NH3: 0.666 atm
To summarize, you calculated the number of moles of CuO, determined the stoichiometry of the reaction to find the number of moles of each gas produced, calculated the total moles of gas in the reaction vessel, and then used the ideal gas law to find the partial pressures of each gas based on the ratio of moles.