Ammonia is formed by reacting nitrogen and hydrogen gases. How many liters of ammonia gas is formed from 13.7 L of hydrogen gas at 93C and pressure of 40 kPa?

If you don't mind do you think you can work the whole thing for me? sorry im just not felling good and im hurting. And i have no clue how to do this stuff. Please?

To solve this problem, we need to use the ideal gas law equation. The equation is given as:

PV = nRT

Where:
P = pressure in kilopascals (kPa)
V = volume in liters (L)
n = number of moles of gas
R = ideal gas constant (0.0821 L·atm/(mol·K))
T = temperature in Kelvin (K)

First, we need to convert the temperature from Celsius to Kelvin by adding 273 to it.

93°C + 273 = 366 K

Now let's calculate the number of moles of hydrogen gas.

Using the ideal gas law equation, we rewrite the equation as:

n = PV / RT

Plugging in the values:
P = 40 kPa
V = 13.7 L
R = 0.0821 L·atm/(mol·K)
T = 366 K

n = (40 kPa * 13.7 L) / (0.0821 L·atm/(mol·K) * 366 K)
n ≈ 6.09 moles of hydrogen gas

According to the balanced chemical equation, the molar ratio between hydrogen gas and ammonia gas is 3:2.
Therefore, for every 3 moles of hydrogen gas, we have 2 moles of ammonia gas.

Thus, the number of moles of ammonia gas formed is (2/3) * 6.09 ≈ 4.06 moles of ammonia gas.

Finally, let's convert moles of ammonia gas into liters using the ideal gas law equation:

V = nRT / P

Plugging in the values:
n = 4.06 moles of ammonia
R = 0.0821 L·atm/(mol·K)
T = 366 K
P = 40 kPa

V = (4.06 moles * 0.0821 L·atm/(mol·K) * 366 K) / 40 kPa
V ≈ 37.2 L

Therefore, approximately 37.2 liters of ammonia gas is formed from 13.7 liters of hydrogen gas at 93°C and a pressure of 40 kPa.

To determine the amount of ammonia gas formed, we need to use the balanced chemical equation for the reaction between nitrogen and hydrogen gases:

N₂(g) + 3H₂(g) → 2NH₃(g)

From the equation, we can see that one mole of nitrogen reacts with three moles of hydrogen to produce two moles of ammonia. We need to convert the given volume of hydrogen gas to moles using the ideal gas law equation:

PV = nRT

Where:
P = pressure (in atm)
V = volume (in liters)
n = moles
R = gas constant (0.0821 L·atm/mol·K)
T = temperature (in Kelvin)

First, we need to convert the temperature from Celsius to Kelvin by adding 273.15:

T = 93°C + 273.15 = 366.15 K

Next, we can rearrange the ideal gas law equation to solve for moles:

n = PV / RT

Substituting the given values:

n = (40 kPa * 13.7 L) / (0.0821 L·atm/mol·K * 366.15 K)

Now, convert the pressure from kilopascals (kPa) to atmospheres (atm) by dividing by 101.325 (since 1 atm = 101.325 kPa):

n = (40 / 101.325) * 13.7 L / (0.0821 L·atm/mol·K * 366.15 K)

Calculate the value of n to find the number of moles of hydrogen gas:

n ≈ 0.5562 moles of H₂

Since the balanced equation tells us that 3 moles of H₂ produce 2 moles of NH₃, we can determine the amount of ammonia gas formed:

(2 moles NH₃ / 3 moles H₂) * 0.5562 moles H₂ = 0.3708 moles of NH₃

Finally, we can convert moles of ammonia to liters using the ideal gas law equation:

V = nRT / P

V = (0.3708 moles * 0.0821 L·atm/mol·K * 366.15 K) / 40 kPa

Again, divide the pressure by 101.325 to convert from kPa to atm:

V = (0.3708 * 0.0821 * 366.15) / (40 / 101.325)

Calculate the value of V to find the volume of ammonia gas:

V ≈ 1.048 L of NH₃

Therefore, approximately 1.048 liters of ammonia gas is formed from 13.7 L of hydrogen gas.

write the balance equation. Determine the moles of H2 (from PV=nRT). Use the coefficents of the balanced equation to determine how many moles of ammonia are made.

I assume you are just starting chemistry, in reality, it is much more complicated that this, as that reaction does not proceed to completion, but rather is a function of equilibrium.