# Consider the reaction: K_2S_((aq))+Co(NO_3)_(2(aq)) -> 2KNO_(3(aq))+CoS_((s)) darr. What volume of 0.220M K_2S solution is required to completely react with 160mL of 0.145M Co(NO_3)_2?

May 20, 2015

You need 105 mL of 0.220-M potassium sulfide solution to completely react with that much cobalt nitrate.

${K}_{2} {S}_{\left(a q\right)} + C o {\left(N {O}_{3}\right)}_{2 \left(a q\right)} \to 2 K N {O}_{\textrm{3 \left(a q\right]}} + C o {S}_{\left(s\right)} \downarrow$

Notice that you have a $1 : 1$ mole ratio between potassium sulfide and cobalt nitrate. This tells you that you need equal numbers of moles of each reactant in order for the reaction to take place.

Figure out how many moles of cobalt nitrate are present in solution by using the solution's molarity and volume

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

${n}_{C o {\left(N {O}_{3}\right)}_{2}} = \text{0.145 M" * 160 * 10^(-3)"L" = "0.0232 moles}$ $C o {\left(N {O}_{3}\right)}_{2}$

You need to have 0.0232 moles of potassium sulfide present in order for the cobalt nitrate to react completely. Once again, use the solution's molarity to determine what volume you'd need

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

V_(K_2S) = (0.0232 cancel("moles"))/(0.220cancel("moles")/"L") = "0.10545 L"

Expressed in mL and rounded to two sig figs, the number of sig figs you gave for the volume of the cobalt nitrate solution, the answer will be

${V}_{{K}_{2} S} = \textcolor{g r e e n}{\text{110 mL}}$