Consider the following reaction: Li_2S(aq) + Co(NO_3)_2 (aq) -> 2LiNO_3(aq) + CoS(s). What volume of 0.160 mol*L^-1 Li_2S, solution is required to completely react with 0.130 mol*L^-1 Co(NO_3)_2?

May 6, 2018

Well the reaction relies on the brick-like solubility of transition-metal sulfides.

Explanation:

We gots...

$L {i}_{2} S \left(a q\right) + C o {\left(N {O}_{3}\right)}_{2} \left(a q\right) \rightarrow C o S \left(s\right) \downarrow + 2 L i N {O}_{3} \left(a q\right)$..

Else we could write the net ionic equation:

$C {o}^{2 +} + {S}^{2 -} \rightarrow C o S \left(s\right) \downarrow$

And so we gots with respect to $C {o}^{2 +}$ $0.130 \cdot m o l \cdot {L}^{-} 1$...

$0.130 \cdot m o l \cdot {L}^{-} 1 \times 100 \cdot m L \times {10}^{-} 3 \cdot L \cdot m {L}^{-} 1 = 0.0130 \cdot m o l$...the volume of the cobalt ion solution was interpolated.

And given the 1:1 stoichiometry....we take the quotient...

$\frac{0.0130 \cdot m o l}{0.160 \cdot m o l \cdot {L}^{-} 1} \times 1000 \cdot m L \cdot {L}^{-} 1 = 81.3 \cdot m L$ of the lithium sulfide solution. Ad please note that I had to interpolate the concentration of lithium sulfide.