posted by caroline on .
A mixture contained zinc sulfide, ZnS, and lead sulfide, PbS. A sample of the mixture weighing 6.12 g was reacted with an excess of hydrochloric acid. The reactions are given below.
ZnS(s) + 2 HCl(aq) → ZnCl2(aq) + H2S(g)
PbS(s) + 2 HCl(aq) → PbCl2(aq) + H2S(g)
If the sample reacted completely and produced 1.038 L of hydrogen sulfide, H2S, at 23°C and 745 mmHg, what were the percentages of ZnS and PbS in the mixture?
Let X = mass ZnS
Let Y = mass PbS
MM = molar mass
You need two equations.
X + Y = 6.12 is one. The second is
X(MM H2S/MM ZnS) + Y(MMH2S/MMPbS) = grams H2S.
The g H2S must be calculated from PV = nRT and solve for n=number of moles H2S, then convert to grams, knowing 22.4 L H2S at STP will have a mass of 34.08 grams.
Solve the two equation simultaneously which will give you grams ZnS and grams PbS. Then percent ZnS = (mass ZnS/mass sample)*100 = ?? and
%PbS = (mass PbS/mass sample)*100 = ??