Question #4858b

1 Answer
Oct 2, 2017

Answer:

#"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.