# Question 12be1

Jun 12, 2016

${\text{87.9 g AgNO}}_{3}$

#### Explanation:

Start by writing the balanced chemical equation that describes this double replacement reaction

color(red)(2)"AgNO"_ (3(aq)) + "FeCl"_ (2(aq)) -> 2"AgCl"_ ((s)) darr + "Fe"("NO"_ 3)_ (2(aq))

Here you have $\textcolor{red}{2}$ moles of silver nitrate, ${\text{AgNO}}_{3}$, reacting with $1$ mole of iron(II) chloride, ${\text{FeCl}}_{2}$, to produce $2$ moles of silver chloride and $1$ mole of aqueous iron(II) nitrate.

You know that the two reactants are used up in a $\textcolor{red}{2} : 1$ mole ratio, which means that you can use their molar masses to convert this into a gram ratio.

You will have

color(red)(2)color(red)(cancel(color(black)("moles AgNO"_3))) * "169.87 g"/(1color(red)(cancel(color(black)("mole AgNO"_3)))) = "339.74 g"

and

1color(red)(cancel(color(black)("mole FeCl"_2))) * "126.75 g"/(1color(red)(cancel(color(black)("mole FeCl"_2)))) = "126.75 g"

Therefore, the $\textcolor{red}{2} : 1$ mole ratio is equivalent to a $339.74 : 126.75$ gram ratio.

Since you know that the reaction must consume $\text{32.8 g}$ of iron(II) chloride, you can say that you need

32.8 color(red)(cancel(color(black)("g FeCl"_2))) * "339.74 g AgNO"_3/(126.75color(red)(cancel(color(black)("g FeCl"_2)))) = color(green)(|bar(ul(color(white)(a/a)color(black)("87.9 g AgNO"_3)color(white)(a/a)|)))#

The answer is rounded to three sig figs.