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

Jun 28, 2017

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

#### Explanation:

We need $\left(i\right)$ a stoichiometric equation........

$3 {I}_{2} \left(a q\right) + N {H}_{3} \left(a q\right) \rightarrow N {I}_{3} \left(s\right) + 3 H I \left(a q\right)$

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

$\left(i i\right)$ $N {I}_{3} \left(s\right) \rightarrow \frac{1}{2} {N}_{2} \left(g\right) + \frac{3}{2} {I}_{2} \left(s\right) + \Delta$

$\text{Moles of nitrogen triiodide} = \frac{3.58 \cdot g}{394.72 \cdot g \cdot m o {l}^{-} 1} = 9.07 \times {10}^{-} 3 \cdot m o l$

And thus we get $\frac{3}{2}$ equiv diiodine......

$\frac{3}{2} \times 9.07 \times {10}^{-} 3 \cdot m o l \times 126.90 \cdot g \cdot m o {l}^{-} 1 \times 2 = 3.45 \cdot g .$

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.