What volume of 0.255 M K2S solution is required to completely react with 170 mL of 0.110 M Co(NO3)2?

##”73 mL”##

Take a look at the balanced chemical equation for this

##”K”_2″S”_text((aq]) + “Co”(“NO”_3)_text(2(aq]) -> 2″KNO”_text(3(aq]) + “CoS”_text((s]) darr##

Notice that you have a ##1:1## between potassium sulfide, ##”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 of the cobalt(II) nitrate. As you know, a solution’s is defined as the number of moles of divided by the volume of the solution – expressed in liters!

##color(blue)(c = 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 = n/V implies n = c * 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 ##”0.255-M”## ##”K”_2″S”## solution will contain ##0.0187## moles of ##”K”_2″S”##

##c = n/V implies V = 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 , 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 = color(green)(“73 mL”)##