what volume of ammonia gas will be produced when 8.01 L of hydrogen react completely in the following equation.
N2 + 3H2 > 2NH3
this stuff is so confusing because one little word seems to change the entire process for solving.
See you other post; the one with 1.73L N2.
It shouldn't be confusing. All of these stoichiometry problems follow the same route.
1. Convert g or L to mols.
2. Use the equation to convert mols of what you have to mols of what you want.
3. Then convert back to the units you want; i.e., g or L.
Sometimes an extra step is inserted because something is different but they are fairly standard.
To find the volume of ammonia gas produced, we first need to determine the stoichiometry of the reaction and then use the volume ratios from the balanced equation.
The balanced equation is:
N2 + 3H2 → 2NH3
From the equation, we can see that 1 mole of N2 reacts with 3 moles of H2 to produce 2 moles of NH3.
Now, we need to convert the given volume of hydrogen gas (H2) to moles. To do this, we will use the ideal gas law:
PV = nRT
Where:
P is the pressure (which we assume to be constant)
V is the volume of the gas (given as 8.01 L)
n is the number of moles of the gas (which we need to find)
R is the ideal gas constant (0.0821 L·atm/(mol·K))
T is the temperature
Since both the pressure and temperature are not specified in the problem, we can disregard them.
Now, rearranging the ideal gas law equation, we get:
n = PV/RT
Substituting the given values:
n = (8.01 L)(1 atm)/(0.0821 L·atm/(mol·K))
n = 97.67 mol
Therefore, we have 97.67 moles of hydrogen gas (H2) reacting.
Using the stoichiometry of the balanced equation, we know that 3 moles of H2 reacts to produce 2 moles of NH3.
So, using the moles of H2, we can calculate the moles of ammonia gas (NH3) produced:
moles of NH3 = (97.67 mol H2)(2 mol NH3 / 3 mol H2)
moles of NH3 = 65.11 mol
Finally, to convert the moles of ammonia gas to volume, we use the ideal gas law again:
V = nRT/P
Given that we assume constant pressure and temperature, we can use the same values for R and P as before.
V = (65.11 mol)(0.0821 L·atm/(mol·K))/(1 atm)
V = 5.34 L
Therefore, approximately 5.34 L of ammonia gas will be produced when 8.01 L of hydrogen reacts completely in the given reaction.