1.09 g #H_2# is allowed to react with 10.2 g #N_2#, producing 1.12 g #NH_3#. What is the theoretical yield in grams for this reaction under the given conditions?

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Answer 1 · 2017-03-07T06:48:22

Less than #20%#..............

We need a stoichiometric equation:

#1/2N_2(g) + 3/2H_2(g)rightleftharpoonsNH_3(g)#

#"Moles of dinitrogen"# #=# #(10.2*g)/(28.02*g*mol^-1)# #=# #0.364*mol#.

#"Moles of dihydrogen"# #=# #(1.09*g)/(2.02*g*mol^-1)# #=# #0.540*mol#.

#"Moles of ammonia"# #=# #(1.12*g)/(17.03*g*mol^-1)# #=# #0.0658*mol#.

Clearly dihydrogen is the limiting reagent, and at most we can make #2/3# equiv ammonia, i.e.

#"% yield"# #-=# #(0.0658*mol)/(2/3xx0.540*mol)xx100%=??%#