Ammonia is often produced by reacting nitrogen gas with hydrogen gas.
How many liters of ammonia gas can be formed from 12.9 L of hydrogen gas at 93.0°C and a pressure of 43.5 kPa?
(Assume an excess amount of nitrogen gas.)
I have absolutely no idea how to begin this.
balance the equation:
3H2+N2>>2NH3
so you get 2/3 of the moles of Hydrogen gas..
moles H2=PV/RT
Use PV=nRT. then substitute all the given and you can find the answer.
To calculate the number of liters of ammonia gas formed, we need to use the ideal gas law equation, which states:
PV = nRT
Where:
P = pressure (in atmospheres)
V = volume (in liters)
n = moles
R = ideal gas constant (0.0821 L•atm/(mol•K))
T = temperature (in Kelvin)
First, we need to convert the given temperature from Celsius to Kelvin. The Celsius to Kelvin conversion is done by adding 273.15 to the given value. So, 93.0°C + 273.15 = 366.15 K.
Now, we can rearrange the ideal gas law equation to solve for moles:
n = PV / RT
Since we are assuming an excess amount of nitrogen gas, we only need to consider the hydrogen gas. The pressure and the temperature are given as 43.5 kPa and 366.15 K, respectively. However, the ideal gas law equation requires pressure to be in atmospheres and temperature to be in Kelvin. So, we convert the pressure from kPa to atm:
43.5 kPa * (1 atm / 101.325 kPa) = 0.4287 atm
Now we can substitute the values into the equation:
n = (0.4287 atm * 12.9 L) / (0.0821 L•atm/(mol•K) * 366.15 K)
Simplifying:
n = (5.5293 L•atm) / (30.0704 L•atm/(mol•K))
n ≈ 0.184 mol
Finally, we need to convert moles of ammonia gas to liters using the molar volume of a gas at standard temperature and pressure (STP), which is 22.4 L/mol:
Volume of ammonia gas = 0.184 mol * 22.4 L/mol
Volume of ammonia gas ≈ 4.114 L
Therefore, approximately 4.114 liters of ammonia gas can be formed from 12.9 L of hydrogen gas.