When a sample of NH3(g) (296.5 grams) is placed in 140.0 L reaction vessel at 676.0 °C and allowed to come to equilibrium the mixture contains 6.250 mol of N2(g). What is the concentration (mol/L) of H2(g)?
2NH3(g) = N2(g)+3H2(g)
See your first post below.
is this simply 3x=6.250
x= 2.083
therefore 2.083/140 = .01489?
I don't think so but that's the right idea.
296.5g NH3/17 = about 17 mols but you can do that more accurately. Actually, you don't need this number at all.
........2NH3 ==> N2 + 3H2
I.......17.44.....0.....0
C......-2x.......x......3x
E.....17.44-x.....x......3x
The problem tells you that N2 = 6.250 mols. That means mols H2 = 3 x 6.250 = ? mols. Then (H2) = mols/L.
To find the concentration of H2(g), we first need to calculate the number of moles of H2(g) in the mixture.
According to the balanced equation:
2NH3(g) = N2(g) + 3H2(g)
We know that the moles of NH3(g) is given as 296.5 grams. To find the moles of NH3(g), we need to divide the given mass by the molar mass of NH3.
The molar mass of NH3(g) can be calculated by summing the atomic masses of its individual atoms:
Molar mass of N = 14.01 g/mol
Molar mass of H = 1.01 g/mol
Molar mass of NH3(g) = (3 * Molar mass of H) + Molar mass of N
= (3 * 1.01 g/mol) + 14.01 g/mol
= 3.03 g/mol + 14.01 g/mol
= 17.04 g/mol
Now let's calculate the moles of NH3(g):
Moles of NH3(g) = Mass of NH3(g) / Molar mass of NH3(g)
= 296.5 g / 17.04 g/mol
≈ 17.37 mol
Since the balanced equation tells us that 2 moles of NH3(g) react to form 3 moles of H2(g), we can calculate the moles of H2(g) as follows:
Moles of H2(g) = (3/2) * Moles of NH3(g)
= (3/2) * 17.37 mol
≈ 26.06 mol
Given that the mixture contains 6.250 mol of N2(g), we can determine that the moles of N2(g) and H2(g) are equal, as given by the balanced equation.
Therefore, the concentration (mol/L) of H2(g) can be calculated by dividing the moles of H2(g) by the volume of the reaction vessel (140.0 L):
Concentration of H2(g) = Moles of H2(g) / Volume of reaction vessel
= 26.06 mol / 140.0 L
≈ 0.186 moles per liter (mol/L)
So, the concentration of H2(g) is approximately 0.186 mol/L.