Nh3

C

cantien98

$N-2$ + 3$H_2$ --> 2$NH_3$
bđ 30......30
pư 0,5x < ----- 1,5x <---------x

ta có H = [TEX]\frac{0,5x}{30}[/TEX]
=> x = 18 l
vậy $NH_3$ = 18l
 
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