Ammonia (NH3) is formed industrially by reacting nitrogen and hydrogen gases. How many liters of ammonia gas can be formed from 13.7L of hydrogen gas at 93.0 degrees Celsius and a pressure of 2.25atm?

9.13 L is correct. By the way, there is a much easier way to do this that I forgot to post after doing it the hard way. But th long way ALWAYS works and this short cut works only in circumstances where all of the materials are gaseous at the conditions listed AND we want the final conditions at th same P and T as the initial conditions. When all are gases, we can cut out the mole step and simply use volumes.

13.7 L x (2 mols NH3/3 mols H2) = 13.7 x 2/3 = 9.13 L. Quick, huh?

Erm.. ok..

PV = nRT
(2.25)(13.7) = n(.08206)(366.15)
n = 1.026

1.026molH2 * 2molNH3/3molH2 = .684molNH3
PV=nRT
(?)(?)=(.684)(.08206)(?)

ok i wrote the other coefficients wrong wow its getting too late for this

Write and balance the equation.

N2 + 3H2 ==> 2NH3.
Use PV = nRT to determine n (# mols) for H2 at the conditions listed.
Convert mols H2 to mols NH3 using the coefficients in the balanced equation.
I assume the question is asking for volume of NH3 produced at the non-standard conditions. Use PV = nRT to convert mols to volume at the non-standard conditions. Post your work if you get stuck.

Ohh, nevermind, your saying I should use the same values I used before for P and T ok..

what I got was 9.13L NH3
thank you so much for your help!

To calculate the number of liters of ammonia gas that can be formed from 13.7L of hydrogen gas, we need to use the balanced chemical equation for the reaction between nitrogen and hydrogen gases to form ammonia:

N2 + 3H2 -> 2NH3

From the equation, we can see that each mole of nitrogen gas reacts with 3 moles of hydrogen gas to form 2 moles of ammonia gas.

To calculate the number of moles of hydrogen gas, we can use the ideal gas law equation:

PV = nRT

Where:
P = Pressure in atm
V = Volume in liters
n = Number of moles
R = Gas constant (0.0821 L.atm/mol.K)
T = Temperature in Kelvin

First, let's convert the temperature from Celsius to Kelvin:
T(K) = T(°C) + 273.15
T(K) = 93.0 + 273.15
T(K) = 366.15 K

Now, let's substitute the values into the equation:
(2.25 atm)(V) = n(0.0821 L.atm/mol.K)(366.15K)

We need to solve for n, the number of moles of hydrogen gas.
n = (2.25 atm)(V) / (0.0821 L.atm/mol.K)(366.15K)

Since we know the volume of hydrogen gas is 13.7L, we can substitute it into the equation:
n = (2.25 atm)(13.7L) / (0.0821 L.atm/mol.K)(366.15K)

Now we have the number of moles of hydrogen gas. According to the balanced chemical equation, 3 moles of hydrogen gas react to form 2 moles of ammonia gas. So, the number of moles of ammonia gas can be calculated by dividing the number of moles of hydrogen gas by 3, and then multiplying it by 2:

n(NH3) = (n(H2) / 3) * 2

Finally, we can convert the number of moles of ammonia gas back into liters using the ideal gas law equation with the same temperature and pressure:

(2.25 atm)(V) = n(NH3)(0.0821 L.atm/mol.K)(366.15K)

We can solve for V, the volume of ammonia gas in liters:
V = (n(NH3)(0.0821 L.atm/mol.K)(366.15K)) / 2.25 atm

By substituting the value of n(NH3) into the equation and solving for V, we can determine the volume of ammonia gas formed from 13.7L of hydrogen gas.

now if i wrote out the equation..

N2 + H2 ==> NH3
then balanced it..
3N2 + H2 ==> 2NH3
could the numbers given possibly just be extra information that is unneeded, and the volume of ammonia simply be 2 * 22.42 because its at STP?
i really would appreciate some help here..