How many moles of #I_2# will form 3.58 g of #NI_3#?

1 Answer
Jun 28, 2017

Answer:

We gets approx. #0.014*mol#........with respect to diiodine......

Explanation:

We need #(i)# a stoichiometric equation........

#3I_2(aq) + NH_3(aq) rarr NI_3(s) + 3HI(aq)#

Which is the synthesis reaction.........and the decomposition reaction is......

#(ii)# #NI_3(s) rarr 1/2N_2(g) +3/2I_2(s)+Delta#

#"Moles of nitrogen triiodide"=(3.58*g)/(394.72*g*mol^-1)=9.07xx10^-3*mol#

And thus we get #3/2# equiv diiodine......

#3/2xx9.07xx10^-3*molxx126.90*g*mol^-1xx2=3.45*g.#

This used to be used as a shock sensitive detonator. When it's dry it's pretty hairy and is to be treated with respect.