# Question b63a9

Dec 7, 2015

$\text{73 mL}$

#### Explanation:

Your strategy here will be to

• write the balanced chemical equation for this double replacement reaction

• examine the mole ratio that exists between the two reactants

• use the molarity and volume of the cobalt(II) nitrate solution to determine how many moles of this compound are present in solution

• use the aforementioned mole ratio to figure out the number of moles of potassium sulfide needed for the reaction

• use the molarity and number of moles of potassium sulfide to find the volume you're looking for

So, the balanced chemical equation for this reaction looks like this

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

Notice the $1 : 1$ mole ratio that exists between the two reactants. This tells you that the reaction consumes equal numbers of moles of potassium sulfide and cobalt(II) nitrate.

As you know, molarity is defined as moles of solute per liters of solution. The number of moles of cobalt(II) nitrate present in that sample solution will be equal to

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

n_(Co(NO_3)_2) = "0.110 M" * 170 * 10^(-3)"L" = "0.0187 moles Co"("NO"_3)_2

In order for the cobalt(II) nitrate to be completely consumed, you need an equal number of moles of potassium sulfide. Therefore,

${n}_{{K}_{2} S} = {n}_{C o {\left(N {O}_{3}\right)}_{2}} = \text{0.0187 moles}$

Finally, use the molarity of the potassium sulfide solution to determine what volume would contain this many moles

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

V_(K_2S) = (0.0187 color(red)(cancel(color(black)("moles"))))/(0.255 color(red)(cancel(color(black)("moles")))/"L") = "0.0733 L"#

Expressed in milliliters and rounded to two sig figs, the answer will be

${V}_{{K}_{2} S} = \textcolor{g r e e n}{\text{73 mL}}$