# Question d6ee2

Jul 15, 2016

${\text{0.40 mol dm}}^{- 3}$

#### Explanation:

The first thing to do here is write the balanced chemical equation that describes this double replacement reaction

2"AgNO"_ (3(aq)) + "MgCl"_ (2(aq)) -> color(blue)(2)"AgCl"_ ((s)) darr + "Mg"("NO"_ 3)_ (2(aq))

Notice that for every mole of magnesium chloride that takes part in the reaction, the reaction consumes $2$ moles of silver nitrate and produces $\textcolor{b l u e}{2}$ moles of silver chloride.

Silver nitrate is in excess, which means that you don't have to worry about this reactant.

Use the molar mass of silver chloride to determine how many moles were produced by the reaction

2.87 color(red)(cancel(color(black)("g"))) * "1 mole AgCl"/(143.32color(red)(cancel(color(black)("g")))) = "0.02003 moles AgCl"

This means that the reaction must have consumed

0.02003 color(red)(cancel(color(black)("moles AgCl"))) * "1 mole MgCl"_2/(color(blue)(2)color(red)(cancel(color(black)("moles AgCl")))) = "0.010015 moles MgCl"2

Now all you have to do is use the volume of the magnesium chloride solution to figure out what concentration of solution would have $0.010015$ moles of magnesium chloride in ${\text{25 cm}}^{3}$ of solution.

Do not forget to convert the volume from cubic centimeters to cubic decimeters by using the conversion factor

$\textcolor{p u r p \le}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{\text{1 dm"^3 = 10^3"cm}}^{3}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

Plug in your values to find

c_("MgCl"_2) = "0.010015 moles"/(25 * 10^(-3)"dm"^3) = color(green)(|bar(ul(color(white)(a/a)color(black)("0.40 mol dm"^(-3))color(white)(a/a)|)))#

The answer is rounded to two sig figs, the number of sig figs you have for the volume of the solution.