Question

Question #4858b

Answer 1

`"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.

In your case, you have

`2"Al"_ ((s)) + 3"I"_ (2(s)) -> 2"AlI"_ (3(s))`

According to the balanced chemical equation, you have

• `"For Al: " "coefficient of 2"`
• `"For I"_2: " coefficient of 3"`
• `"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 color(red)(cancel(color(black)("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.