Question

mttps://www.o... Student Cons... Diagnostic Re... Quiz: E D Question 55 1 pts 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: N2O4 = 92.02 g/mol, N H4 = 32.05 g/mol. N2O4(1) + 2 N2H4(1) 3 N2(g) + 4H2O(g) OLR-N2H4, 59.0 g N formed OLR-N204, 105 g Ny formed No LR, 45.0 g N2formed OLR-N2H4, 13.3 g N formed OLR-N204, 45.7 g Ny formed

Answer 1

mol = weight molecularmass mol at N2o y = 50 = 0.543 mol 92.02 mol at Na Hy = 45 = 1.40 mol 32.05 N2ou (l) + 2N2 Huld) 3N,9 +4 12019) intial 0.543 1.40 0 Mol 0 Limiting reagent 0.543 -0.543 1.40 = 0.70 T = Intial mol Stichiometry = L.R. 10.543) 0.73 0.73 So. N2 Ou is limiting reageant.

Nroy + 2N2 Hu 1 mol at N₂ Ou form 3N2 + 4H2O 3 mol at N2 I molat Nou - 3 mol at N2 0.543 mol af Nou = 0,543x3 molaf No = 1,629 mor at Nr som during So. 0.543x3 the 8th. in 8th 1.629 molat No Present. all son mol = weight moleculwl mass molecular mass at W2 = 28. So. 1.629 428 - weight weight at Wris - 45.612 in 45.79 So. L. R. = N2ou, 45.79 N2 formed.