What volume of nitrogen at 215 degress celcius and 715 mmHg would be required to react with .100 moles of hydrogen to produce ammonia?

N2+3H2--> 2NH3

To calculate the volume of nitrogen required for the given reaction, we need to use the ideal gas law equation:

PV = nRT

Where:
P = pressure of the gas (in this case, nitrogen)
V = volume of the gas (to be determined)
n = number of moles of the gas (in this case, nitrogen)
R = ideal gas constant (0.0821 L·atm/(mol·K))
T = temperature of the gas (in Kelvin)

First, let's convert the given temperature from Celsius to Kelvin:
T(K) = 215°C + 273.15

T(K) = 488.15 K

Next, we'll rearrange the ideal gas law equation to solve for V:
V = (nRT) / P

Given:
n = 0.100 moles (the stoichiometric coefficient for nitrogen is 1, so we need the same number of moles of nitrogen as hydrogen)
P = 715 mmHg

Let's convert the pressure from mmHg to atm:
P(atm) = 715 mmHg / 760 mmHg/atm

P(atm) = 0.940 atm

Now, we can substitute the given values into the equation to calculate the volume of nitrogen:
V = (0.100 moles * 0.0821 L·atm/(mol·K) * 488.15 K) / 0.940 atm

V ≈ 4.278 L

Therefore, approximately 4.278 liters of nitrogen at 215 degrees Celsius and 715 mmHg would be required to react with 0.100 moles of hydrogen to produce ammonia.