Asked by Lisa
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?
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?
Answers
Answered by
DrBob222
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).
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).
Answered by
Lisa
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.