How many moles of $I_{2}$ $I_2$ will form 3.58 g of $N I_{3}$ $NI_3$ ?

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Answer 1 · 2017-06-28T12:38:31

We gets approx. $0.014 \cdot m o l$ $0.014*mol$ ........with respect to diiodine......

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

$3 I_{2} (a q) + N H_{3} (a q) \to N I_{3} (s) + 3 H I (a q)$ $3I_2(aq) + NH_3(aq) rarr NI_3(s) + 3HI(aq)$

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

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

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

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.

How many moles of I_2I2I_2 will form 3.58 g of NI_3NI3NI_3?