Answer : The number of moles of [tex]I_2[/tex] form will be, 0.01360 moles
Solution : Given,
Mass of [tex]NI_3[/tex] = 3.58 g
Molar mass of [tex]NI_3[/tex] = 394.71 g/mole
Molar mass of [tex]I_2[/tex] = 253.80 g/mole
First we to calculate the moles of moles of [tex]NI_3[/tex].
[tex]\text{Moles of }NI_3=\frac{\text{Mass of }NI_3}{\text{Molar mass of }NI_3}=\frac{3.58g}{394.71g/mole}=9.069\times 10^{-3}moles[/tex]
Now we have to calculate the moles of [tex]I_2[/tex].
The balanced reaction is,
[tex]2NI_3\rightarrow N_2+3I_2[/tex]
From the balanced reaction, we conclude that
As, 2 moles of [tex]NI_3[/tex] gives 3 moles of [tex]I_2[/tex]
So, [tex]9.069\times 10^{-3}[/tex] moles of [tex]NI_3[/tex] gives [tex]\frac{3}{2}\times (9.069\times 10^{-3})=0.01360[/tex] moles of [tex]I_2[/tex]
Therefore, the number of moles of [tex]I_2[/tex] form will be, 0.01360 moles
