Respuesta :

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]

  • There is 2moles of Ammonia
  • 3moles of H_2
  • 1mole of N_2

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]

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