A 453 K and 755 mm Hg , what volume of N2 completely react with 22.2 L H2 to produce NH3 ?
N2 ( g ) + 3 H2 ( g ) -> 2 NH3 ( g )
I doubt if I use the ideal gas formula PV = nRT or another , but having only a pressure and a temperature which is not used because it is not what should be the value of my constant n
To solve this problem, you can use the ideal gas law formula, PV = nRT, where:
P is the pressure in units of pressure (mm Hg in this case)
V is the volume in units of liters (L)
n is the number of moles of the gas (N2 or NH3)
R is the ideal gas constant with a value of 0.0821 L·atm/(mol·K)
T is the temperature in units of Kelvin (K)
Since you have only the pressure and volume for N2, and you want to find the volume of N2 that reacts with a known volume of H2, you can start by calculating the number of moles of H2 and then use the balanced equation to determine the number of moles of N2.
First, let's convert the pressure from mm Hg to atm by dividing it by 760 mm Hg/1 atm:
P (atm) = 755 mm Hg / 760 mm Hg/atm = 0.9934 atm
Next, you need to calculate the number of moles of H2 using the ideal gas law formula:
PV = nRT
n(H2) = PV / RT
Plugging in the known values, we get:
n(H2) = (0.9934 atm) (22.2 L) / (0.0821 L·atm/(mol·K)) (453 K)
n(H2) ≈ 1.0892 moles
According to the balanced equation, 1 mole of N2 reacts with 3 moles of H2 to produce 2 moles of NH3. Therefore, the number of moles of N2 needed would be:
n(N2) = (1.0892 moles H2) / (3 moles N2)
n(N2) ≈ 0.3631 moles
Now, you can use the ideal gas law formula once again to calculate the volume of N2:
PV = nRT
(0.9934 atm) (V) = (0.3631 moles) (0.0821 L·atm/(mol·K)) (453 K)
V ≈ (0.3631 moles) (0.0821 L·atm/(mol·K)) (453 K) / 0.9934 atm
V ≈ 13.19 L
Therefore, approximately 13.19 L of N2 would completely react with 22.2 L of H2 to produce NH3.