# 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.

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.