How many moles of hydrogen gas are required to produce 75.0 g of ammonia?

Question

1 Ng + 3 H2 ----------->2 NH3

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Answer 1 · 2018-07-11T05:36:17

We interrogate the reaction...

#1/2N_2(g) + 3/2H_2(g) stackrel("catalysis")rarr NH_3(g)#

We require #(75.0*g)/(17.03*g*mol^-1)=4.40*mol#..with respect to AMMONIA.

And so we require #3/2*"equiv"# of dihydrogen gas...and this equivalence I read directly from the stoichiometrically balanced equation....

i.e. #3/2xx4.40*mol-=6.61*mol#...i.e. a mass of approx...#13.2*g#...with RESPECT TO DIHYDROGEN....