What volume of NO measured at 1.1 atm and 1100 C, can be produced from 10.0 L of NH3 and excess O2 measured at the same temperature and pressure?
Here is the balance equation
4NH3+3O2+4NO+6H20
So far this is what I have:
(1.1 atm)(10.0L)
__________________=.097mol NH3
.08201 x 1373 K
A mixture of 2.50g H2 and 1.50g He exerts a pressure of 0.780atm. What is the partial pressure of H2 gas present in the mixture?
I think i convert into moles
2.50H2 x 1 mol/2.02g =1.24
1.50He x 1 mol/4.00g= 0.625
You can't have an equation without an arrow and an arrow is necessary for us to know which are the reactants and which are the products.
For the NH3 problem, you have correctly determined moles NH3.
moles NO = 0.0976 x (4 moles NO/4 moles NH3) = ?? Then plug that number back into pV= nRT at same T and P and I would be surprised if you didn't get 10.0 L. Does that give a hint to the short cut you could use?
2. The 1.24 looks ok but you should recalculate the moles He. n is not equal to your number.
Convert moles you have (or will redo for He) to mole fraction, then PH2 = total pressure x mole fraction H2 gas.
To calculate the volume of NO that can be produced in the given reaction, we need to first find the moles of NH3.
Using the ideal gas law equation, PV=nRT, we can rearrange it to solve for moles:
n = (PV) / (RT)
Where:
P = pressure in atm
V = volume in liters
n = moles
R = gas constant (0.0821 L.atm / mol.K)
T = temperature in Kelvin
Given:
Pressure (P) = 1.1 atm
Volume (V) = 10.0 L
Temperature (T) = 1100°C = 1373 K
Now we can substitute these values into the equation to find the moles of NH3:
n = (1.1 atm * 10.0 L) / (0.0821 L.atm / mol.K * 1373 K)
n = 0.097 mol NH3
Now, looking at the balanced equation:
4NH3 + 3O2 → 4NO + 6H2O
We can see that for every 4 moles of NH3, we get 4 moles of NO. Therefore, the moles of NO produced will be equal to the moles of NH3.
So, the volume of NO produced will also be:
(0.097 mol NO) * (22.4 L/mol) = 2.17 L
Therefore, the volume of NO measured at 1.1 atm and 1100°C, produced from 10.0 L of NH3 and excess O2, will be 2.17 L.
Moving on to the second question:
To find the partial pressure of H2 gas present in the mixture, we need to first find the total moles of the gases.
Given:
Mass of H2 = 2.50 g
Molar mass of H2 = 2.02 g/mol
Moles of H2 = (mass of H2) / (molar mass of H2)
Moles of H2 = 2.50 g / 2.02 g/mol = 1.24 mol
Similarly,
Mass of He = 1.50 g
Molar mass of He = 4.00 g/mol
Moles of He = (mass of He) / (molar mass of He)
Moles of He = 1.50 g / 4.00 g/mol = 0.375 mol
Now, to find the total moles of the gases, add the moles of H2 and He:
Total moles = Moles of H2 + Moles of He
Total moles = 1.24 mol + 0.375 mol = 1.615 mol
Given:
Total pressure (P) = 0.780 atm
Now we can find the partial pressure of H2:
Partial pressure of H2 = (moles of H2 / total moles) * total pressure
Partial pressure of H2 = (1.24 mol / 1.615 mol) * 0.780 atm
Partial pressure of H2 = 0.605 atm
Therefore, the partial pressure of H2 gas present in the mixture is 0.605 atm.