During studies of the reaction below, 2 N2H4(l) + N2O4(l) ? 3 N2(g) + 4 H2O(g) a

Question

During studies of the reaction below, 2 N2H4(l) + N2O4(l) ? 3 N2(g) + 4 H2O(g) a chemical engineer measured a less-than-expected yield of N2 and discovered that the following side reaction occurs. N2H4(l) + 2 N2O4(l) ? 6 NO(g) + 2 H2O(g) In one experiment, 11.5 g of NO formed when 102.1 g of each reactant was used. What is the highest percent yield of N2 that can be expected?

Explanation / Answer

PLEASE RATE ME AND AWARD MEKARMA POINTS IF IT IS HELPFUL FOR YOU first of all determine which is the limiting reagent H2 = 2g/mol 1.79g/2g/mol = 0.895 mol N2 = 28 g/mol 9.98g / 28 g/mol = 0.356428 mol => H2 is the limiting reagent as to react all of the N2 you would require 1.069 mol of H2 so now determine the max no. of mol possible = mol H2/3*2 from equation = 0.59666mol NH3 =17g/mol possible weight of NH3 = 0.59666mol*17g/mol =10.143333 g 1.63/10.143333*100/1=16.06%

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