# Question 554f2

Jun 21, 2016

${\text{21 moles PbCl}}_{2}$

#### Explanation:

The coefficients that appear in a balanced chemical equation tell you the mole ratios in which the chemical species that take part in the reaction find themselves.

In this case, the balanced chemical equation that describes this double replacement reaction looks like this

3"Pb"("NO"_ 3)_ (2(aq)) + color(red)(2)"AlCl"_ (3(aq)) -> color(blue)(3)"PbCl"_ (2(s)) darr + 2"Al"("NO"_ 3)_ (2(aq))

You're interested in finding out how many moles of lead(II) chloride, ${\text{PbCl}}_{2}$, are produced when $14$ moles of aluminium chloride, ${\text{AlCl}}_{3}$, react completely.

Notice the $\textcolor{red}{2} : \textcolor{b l u e}{3}$ mole ratio that exists between the two compounds.

This mole ratio tells you that for every $\textcolor{red}{2}$ moles of aluminium chloride that take part in the reaction you get $\textcolor{b l u e}{3}$ moles of lead(II) chloride, provided of course that you have enough of the other reactant to ensure that all the moles of aluminium chloride react.

In your case, the reaction is said to consume $14$ moles of aluminium chloride, which means that it produced

14 color(red)(cancel(color(black)("moles AlCl"_3))) * (color(blue)(3)color(white)(a)"moles PbCl"_2)/(color(red)(2)color(red)(cancel(color(black)("moles AlCl"_3)))) = color(green)(|bar(ul(color(white)(a/a)color(black)("21 moles PbCl"_2)color(white)(a/a)|)))#