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A mixture contained zinc sulfide, ZnS, and lead sulfide, PbS. A sample of the mixture weighing 6.13 g was reacted with an excess of hydrochloric acid. The reactions are




ZnS(s) + 2HCl(aq) ==> ZnCl2(aq) + H2S(g)


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




If the sample reacted completely and produced 1.048 L of hydrogen sulfide, H2S, at 21C and 742 mmHg, what were the percentages of ZnS and PbS in the mixture. Enter the percentage for ZnS first, then PbS.

2006-10-13 05:13:51 · 3 answers · asked by thewhite_stag 1 in Science & Mathematics Chemistry

3 answers

a gram of ZnS produces more H2S than a gram of PbS because PbS weighs much more per mole.

First, figure out how much H2S was produced: PV = nRT means n=PV/RT = 742mmHg*1.048L/294K/62.3637 L · mmHg · K-1 · mol-1 = .0424 mol

Moles of ZnS or moles of PbS = moles of H2S produced, so

x mol ZnS + (.0424-x) mol PbS = 6.13 g
x mol ZnS * 97.43 g/mol ZnS + (.0424-x) * 239.25g/mol PbS = 6.13 g

Now let's do some Algebra.

97.43 x + 10.14 - 239.25 x = 6.13
-141.82 x = -4.01
x = .0283 moles = number of moles of ZnS

That's 2.76 g. Out of the original sample, that's 2.76 / 6.13 = 45%. The rest, 55%, is PbS.

Check: Number of moles of PbS = .0424 - .0283 = .0141 moles

That's 3.37 g. 3.37/6.13 = 55%. Checks out.

2006-10-13 05:58:26 · answer #1 · answered by Mr. E 5 · 0 0

because of the fact the smelling stuff is acetic acid, an acid/base extraction would do away with it. you would be able to desire to dissolve the blend in an organic and organic solvent and shake it up with a answer of baking soda. this would generate sodium acetate, which might partition into the water layer. Acidification of the water layer would regenerate the acid.

2016-12-26 18:17:58 · answer #2 · answered by jamila 3 · 0 0

As I said.................

2006-10-13 06:57:32 · answer #3 · answered by ag_iitkgp 7 · 0 0

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