# Question 3ab0e

Nov 10, 2015

$\text{73 mL}$

#### Explanation:

Take a look at the balanced chemical equation for this double replacement reaction

${\text{K"_2"S"_text((aq]) + "Co"("NO"_3)_text(2(aq]) -> 2"KNO"_text(3(aq]) + "CoS}}_{\textrm{\left(s\right]}} \downarrow$

Notice that you have a $1 : 1$ mole ratio between potassium sulfide, $\text{K"_2"S}$, and cobalt(II) nitrate, "Co"("NO"_3)_2.
This tells you that the reaction wil consume equal numbers of moles of the two reactants.

Now, you are given the volume and molarity of the cobalt(II) nitrate. As you know, a solution's molarity is defined as the number of moles of solute divided by the volume of the solution - expressed in liters!

$\textcolor{b l u e}{c = \frac{n}{V}}$

This means that you can rearrange the above equation and solve for $n$, the number of moles of cobalt(II) nitrate you have in that solution - do not forget to convert the volume from mililiters to liters

$c = \frac{n}{V} \implies n = c \cdot V$

n = "0.110 M" * 170 * 10^(-3)"L" = "0.0187 moles Co"("NO"_3)_2

Since you've established that you need equal numbers of moles of cobalt(II) nitrate and potassium sulfide, all you need to do now is figure out what volume of the $\text{0.255-M}$ $\text{K"_2"S}$ solution will contain $0.0187$ moles of $\text{K"_2"S}$

$c = \frac{n}{V} \implies V = \frac{n}{c}$

V = (0.0187color(red)(cancel(color(black)("moles"))))/(0.255color(red)(cancel(color(black)("moles")))/"L") = "0.07333 L"#

You need to round the answer to two sig figs, the number of sig figs you have for the volume of the cobalt(II) nitrate solution, and expressed in mililiters, the answer will be

$V = \textcolor{g r e e n}{\text{73 mL}}$