Consider the reaction: K2S(aq)+Co(NO3)2(aq)→ 2KNO3(aq)+CoS(s) What volume of 0.255 M K2S solution is required to completely react with 170 mL of 0.110 M Co(NO3)2? Thank you!
##”73 mL”##
Your strategy here will be to
write the balanced chemical equation for this
examine the that exists between the two reactants
use the 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
##”K”_2″S”_text((aq]) + “Co”(“NO”_3)_text(2(aq]) -> 2″KNO”_text(3(aq]) + “CoS”_text((s]) darr##
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 per liters of solution. The number of moles of cobalt(II) nitrate present in that sample solution will be equal to
##color(blue)(c = n/V implies n = c * 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_2S) = n_(Co(NO_3)_2) = “0.0187 moles”##
Finally, use the molarity of the potassium sulfide solution to determine what volume would contain this many moles
##color(blue)(c = n/V implies V = 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 , the answer will be
##V_(K_2S) = color(green)(“73 mL”)##