How many liters of ammonia gas can be formed from 11.3 L of hydrogen gas at 93.0 C and a pressure of 42.4 kPa?

N2 + 3H2 ==> 2NH3

11.3L x (2 mols NH3/3 mols H2) = 11.3 x 2/3 = L NH3 at STP. Convert to 93 C and 42.4 kPa.

To determine the number of liters of ammonia gas formed, we need to use the balanced chemical equation for the reaction between hydrogen gas (H2) and nitrogen gas (N2) to form ammonia gas (NH3):

3H2 + N2 -> 2NH3

From the balanced equation, we can see that it takes 3 moles of hydrogen gas to produce 2 moles of ammonia gas.

To find the moles of hydrogen gas, we will use the ideal gas law equation:

PV = nRT

Where:
P = pressure (in atm or Pa)
V = volume (in liters or m^3)
n = number of moles
R = gas constant (0.0821 L·atm/(mol·K) or 8.314 J/(mol·K))
T = temperature (in Kelvin)

First, we need to convert the given pressure of 42.4 kPa to atm by dividing it by 101.325 kPa/atm:

42.4 kPa / 101.325 kPa/atm = 0.418 atm

Next, we need to convert the given volume of hydrogen gas from 11.3 L to moles. To do this, we divide the volume by the molar volume of an ideal gas at Standard Temperature and Pressure (STP), which is 22.4 L/mol:

11.3 L / 22.4 L/mol = 0.504 mol

Now, we can determine the moles of ammonia gas formed by using the stoichiometry of the balanced equation. Since it takes 3 moles of hydrogen gas to produce 2 moles of ammonia gas:

2/3 * 0.504 mol = 0.336 mol

Finally, we can convert the moles of ammonia gas to liters using the molar volume at STP:

0.336 mol * 22.4 L/mol = 7.52 L

Therefore, approximately 7.52 liters of ammonia gas can be formed from 11.3 L of hydrogen gas at 93.0 °C and a pressure of 42.4 kPa.