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)
3 answers
See your first post below.
is this simply 3x=6.250
x= 2.083
therefore 2.083/140 = .01489?
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.
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.
2 answers