Solution Found!
During studies of the reaction in Sample 3.13, a chemical
Chapter 3, Problem 84P(choose chapter or problem)
During studies of the reaction in Sample Problem \(3.11\),
\(2 \mathrm{~N}_{2} \mathrm{H}_{4}(\mathrm{l})+\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{l}) \rightarrow 3 \mathrm{~N}_{2}(\mathrm{~g})+4 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\)
Equation Transcription:
2N2H4(l) + N2O4(l) 3N2(g) + 4H2O(g)
Text Transcription:
3.11
2N_2H_4(l) + N_2O_4(l) rightarrow 3N_2(g) + 4H_2O(g)
Questions & Answers
QUESTION:
During studies of the reaction in Sample Problem \(3.11\),
\(2 \mathrm{~N}_{2} \mathrm{H}_{4}(\mathrm{l})+\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{l}) \rightarrow 3 \mathrm{~N}_{2}(\mathrm{~g})+4 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\)
Equation Transcription:
2N2H4(l) + N2O4(l) 3N2(g) + 4H2O(g)
Text Transcription:
3.11
2N_2H_4(l) + N_2O_4(l) rightarrow 3N_2(g) + 4H_2O(g)
ANSWER:
Solution 84P
Here we have to calculate the highest percent yield of N2.
Step 1
The given chemical reactions are,
2 N2H4 (l ) + N2O4 (l) → 3N2 (g) + 4H2O (g)
N2H4 (l ) + 2N2O4 (l) → 6NO (g) + 2H2O (g)
Mass of NO formed = 10.0 g
Mass of each reactant = 100.0 g
1st we have to determine the limiting reactant:-
N2 formed from N2O4 :-
In the above reaction 3 mole of N2 has formed from 1 moles of N2O4
100.0 g N2O4
= 100.0 g N2O4
= 3.26016 mol N2
N2 formed from N2H4 :-
In the above reaction 3 mole of N2 has formed from 1 moles of N2O4
100.0 g N2H4
= 100.0 g N2H4
=4.68019 mol N2
As N2O4 has formed less mole of N2 thus it is the limiting agent.