# Question #4858b

Oct 2, 2017

${\text{0.644 moles I}}_{2}$

#### Explanation:

The trick is to use the coefficients added in front of the chemical species that take part in the reaction in the balanced chemical equation.

$2 {\text{Al"_ ((s)) + 3"I"_ (2(s)) -> 2"AlI}}_{3 \left(s\right)}$

According to the balanced chemical equation, you have

• $\text{For Al: " "coefficient of 2}$
• $\text{For I"_2: " coefficient of 3}$
• $\text{For AlI"_3: " coefficient of 2}$

This tells you that for every $2$ moles of aluminium that take part in the reaction, the reaction consumes $2$ moles of iodine and produces $2$ moles aluminium iodide.

Notice that the two reactants react in a $2 : 3$ mole ratio. You can use this mole ratio to find the number of moles of iodine needed in order to ensure that $0.429$ moles of aluminium react.

$0.429 \textcolor{red}{\cancel{\textcolor{b l a c k}{{\text{moles Al"))) * overbrace("3 moles I"_2/(2color(red)(cancel(color(black)("moles Al")))))^(color(blue)("given by the balanced chemical equation")) = color(darkgreen)(ul(color(black)("0.644 moles I}}_{2}}}}$

The answer is rounded to three sig figs, the number of sig figs you have for the number of moles of aluminium.