[tex]2N_2 + 3H_2 =\ \textgreater \ 2NH_3[/tex]
v amoniac= 2.24 l
n amoniac=2.24 l / 22.4 l/mol = 0.1 mol
3 mol H_{2} .... 2 mol amoniac
a mol H_{2} ....0.1 mol amoniac
a= 0.15 mol H_{2}
nr atomi H = 0.15 mol × [tex]6.022 x 10^{23} atomi/mol[/tex]
= 9.033 x [tex] 10^{22} [/tex]