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?
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).
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.
To determine the composition in moles of each substance in the reaction vessel when the reaction is completed, we need to use stoichiometry.
In this reaction, we have 4 moles of NH3 reacting with 1 mole of N2 to produce 3 moles of N2H4.
Given that 0.250 mol of NH3 and 0.100 mol of N2 were used, we can calculate the moles of N2H4 formed as follows:
From the given molar ratios in the balanced equation, we know that for every 4 moles of NH3 reacted, 3 moles of N2H4 are formed. Therefore, we can first calculate the moles of N2H4 formed from the moles of NH3:
(0.250 mol NH3) x (3 mol N2H4 / 4 mol NH3) = 0.188 mol N2H4
Next, we can calculate the moles of N2H4 formed from the moles of N2:
(0.100 mol N2) x (3 mol N2H4 / 1 mol N2) = 0.300 mol N2H4
So, the composition in moles in the vessel when the reaction is completed is:
- 0.000 mol NH3 (since all the NH3 is consumed)
- 0.100 mol N2 (since N2 is not fully consumed)
- 0.188 mol N2H4 (formed during the reaction)
Therefore, the answer you provided is incorrect. The correct composition is 0.000 mol NH3, 0.100 mol N2, and 0.188 mol N2H4.