Molar mass of NH_3
[tex]\\ \sf\longmapsto 14u+3(1u)[/tex]
[tex]\\ \sf\longmapsto 14u+3u[/tex]
[tex]\\ \sf\longmapsto 17g/mol[/tex]
We know.
No of moles=Given mass/Molar mass
[tex]\\ \sf\longmapsto Given\;Mass=17(5)[/tex]
[tex]\\ \sf\longmapsto Given \:Mass\:of\:NH_3=85g[/tex]
Now
Lets write the balanced equation
[tex]\\ \sf\longmapsto N_2+3H_2=2NH_3[/tex]
Now
[tex]\boxed{\sf No\:of\:Molecules =No\:of\;moles\times Avagadro\:no}[/tex]
For Hydrogen
[tex]\\ \sf\longmapsto 3\times 6.023\times 10^{23}[/tex]
[tex]\\ \sf\longmapsto 18.069\times 10^{23}[/tex]
[tex]\\ \sf\longmapsto 1.8\times 10^{22}molecules[/tex]
For Ammonia
[tex]\\ \sf\longmapsto 2\times 6.023\times 10^{23}[/tex]
[tex]\\ \sf\longmapsto 12.046\times 10^{23}[/tex]
[tex]\\ \sf\longmapsto 1.2\times 10^{22}molecules[/tex]
For Nitrogen
[tex]\\ \sf\longmapsto 1\times 6.023\times 10^{23}[/tex]
[tex]\\ \sf\longmapsto 6.023\times 10^{23}molecules[/tex]