12-A mixture contained zinc sulfide, ZnS, and lead sulfide, PbS. Asample of the

Question

12-A mixture contained zinc sulfide, ZnS, and lead sulfide, PbS. Asample of the mixture weighing 6.12 g was reacted with an excess ofhydrochloric 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 1.048 L of hydrogen sulfide, H2S, at23°C and 745 mmHg, what were the percentages of ZnS and PbS inthe mixture?
ZnS
1%
PbS
2%
ZnS(s) + 2 HCl(aq) ZnCl2(aq) + H2S(g)

PbS(s) + 2 HCl(aq) PbCl2(aq) + H2S(g)

Explanation / Answer

We Know that :    The given equations are :     ZnS(s) + 2 HCl(aq) ZnCl2(aq) +H2S(g)     PbS(s) + 2 HCl(aq) PbCl2(aq) +H2S(g)         According to idealgas equation :        PV = nRT         n ( total) = 745 / 760 atm x 1.048 L / 0.0821 atm-L/ mol-K x 296 K                         = 0.0423 mol         wt. ZnS x M.wtof PbS + wt. PbS x M.wt of ZnS = 0.0423 x M.wt ZnS x M.wt PbS          239.26 wt. ZnS + 97.47 wt. PbS = 986.464                 wt. ZnS + wt. PbS = 6.12          solving theabove linear equations we get :           wt. ZnS   = 2.75 g           wt.PbS = 6.12 - 2.75 g                        = 3.37 g         The % of ZnS in themixture = 2.75 / 6.12 * 100                                                     = 45 %           The %of PbS in the mixture =  3.37 / 6.12 * 100                                                     = 55 %                                                     = 55 %      

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