Question #4858b

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.