Răspuns :
N2 + 3H2 => 2NH3
3*22,4L H2.........2*17g NH3
4,48L H2............xg
x=34*4,48/67,2=
3*22,4L H2.........2*17g NH3
4,48L H2............xg
x=34*4,48/67,2=
N2 + 3H2 ---> 2 NH3
Nr moli H2 =V/V0 =4.48/22.4 = 0.2 moli H2
3 moli H2 ... 2 moli NH3... 2×17=34 g NH3
0.2 moli H2....................................x g NH3
=> x=34×0.2/3
x= 2.2666 g NH3