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QUESTION

# 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")

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