Hey this is my question:
At a temperature of 300°C and a pressure of 40.5 MPa, 90.0 mol of H2(g) and
80.0 mol of N2(g) are injected into a reaction vessel. When equilibrium is
established, 37.0 mol of NH3(g) are present. The number of moles of H2(g) present
in this equilibrium mixture is
This is my equation so far. How do I go from here?
N2(g) + 3H2(g) >> 2 NH3(g)
initially: 80.0 mol 90.0 mol 0
at equilib:80.0 – x 90.0 – 3x 0 + 2x
0+2x=37.0 mol given
So solve for x and use for 90-3x = ??
ok, I got 20 as the H2 (g) present, would this be right?
Hardly.
If 2x = 37, then x = 37/2 =18.5
Then 90-3x=90-3(18.5)=90-55.5 = 34.5. Check my thinking. Check my math.
mmmmmmmmmmm the answer is 45+897x99=11435%19=897
Set up your RICE or ICE table;
(R) 3H2(g) + N2(g) <-> 2NH3(g)
(I) 90 80 0
(C) 90-3x 80-x +2x
(E) 90-3x 80-x 20
Looking at NH3(g) you can calculate that x = 10, therefore 90-3(10)=60. 60M or (60mol/L) is the answer. I hope that formatting sticks and this makes sense.
Let's correct the calculation.
Given the equation: N2(g) + 3H2(g) >> 2 NH3(g)
From the information given:
- Initially: 80.0 mol of N2(g) and 90.0 mol of H2(g) are present.
- At equilibrium: 37.0 mol of NH3(g) are present.
To find the number of moles of H2(g) present at equilibrium, we can use the stoichiometric ratio between H2 and NH3 from the balanced equation.
From the balanced equation, we know that 1 mole of N2 reacts with 3 moles of H2 to produce 2 moles of NH3. Therefore, the stoichiometric ratio between H2 and NH3 is 3:2.
Let's solve for x:
37.0 mol NH3(g) * (3 mol H2(g) / 2 mol NH3(g)) = 55.5 mol H2(g)
So, at equilibrium, the number of moles of H2(g) present is 55.5 mol, not 20 mol.
I'm sorry, but your response doesn't seem to be related to the question you asked earlier. To find the number of moles of H2(g) present in the equilibrium mixture, you need to solve for x in the equation:
2x = 37.0 mol
Divide both sides by 2:
x = 37.0 mol / 2
x = 18.5 mol
Then, substitute the value of x in the expression:
90.0 mol - 3x
90.0 mol - 3(18.5 mol)
90.0 mol - 55.5 mol
34.5 mol
Therefore, the number of moles of H2(g) present in this equilibrium mixture is 34.5 mol.