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
2 answers
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.
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.
3 answers