N2O4(l) + 2N2H4(l) ---------------> 3N2(g) + 4H2O(g) no of moles of N2O4 = W/G.M
ID: 699625 • Letter: N
Question
N2O4(l) + 2N2H4(l) ---------------> 3N2(g) + 4H2O(g)
no of moles of N2O4 = W/G.M.Wt
= 50/92.02 = 0.543 moles
no of moles of N2H4 = W/G.M.Wt
= 45/32.05 = 1.4 moles
2 moles of N2H4 react with 1 moles of N2O4
1.4 moles of N2H4 react with = 1*1.4/2 = 0.7 moles of N2O4 is required
N2O4 is limiting reactant
1 moles of N2O4 react with N2H4 to gives 3 moles of N2
0.54 moles of N2O4 react with N2H4 to gives = 3*0.543/1 = 1.629 moles of N2
mass of N2 = no of moles * gram molar mass
= 1.629*28 = 45.7g of N2
A. LR = N2O4 , 45.7g of N2 formed
Explanation / Answer
11) Determine the limiting reactant (LR) and the mass (in g) of nitrogen that can be formed from 50.0 g N204 and 45.0 g N2H4. Some possibly useful molar masses are as follows: N204-92.02 gmol, N2H4 32.05 g/mol. N2O4+2 N2H4 3 N 2(g) + 4 H2O(g) A) LR = N204, 457 g N2 formed B) LR = N2H4, 13.3 g N2 formed LR = N204, 105 g N2 formed D) LR = N2H4, 59.0 g N2 formed E) No LR, 45.0 g N2 formed
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