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).
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?
2 answers
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.
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.
0 answers