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In a reaction vessel, the following reaction was carried out using 0.250 mol of NH3 and 0.100 mol of N2,
4NH3(l) + N2 (g) = 3N2H4 (l)
what is the compositon in moles in the vessel when the reaction is completed? Is the answer:
0 mol NH3, 0.063 mol N2, and 0.188 mol N2H4?

  • chemistry - ,

    close but not quite.
    Yes, zero moles NH3.
    Yes, 0.1875 = 0.188 mole N2H4.
    No, N2. You USE 0.0625 moles N2, you had 0.100 initially; therefore, the amount remaining is 0.100 - 0.0625 (you are allowed three significant figures so I would not round that to 0.063).

  • chemistry - ,

    Dr. Bob,
    Thank you so much for taking the time to help me. God bless all your work with us. However, I'm still confused b/c these were my choices:

    0 mol NH3, 0.038 mol N2, 0.333 mol N2H4,
    O.150 mol NH3,~0 mol N2, 0.300 mol N2H4,
    ~0 mol NH3, 0.38 mol N2, 0.188 mol N2H4 and
    ~ 0 mol NH3, 0.063 mol N2, 0.188 mol N2H4.

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