Question

Stainless steel is an alloy of iron, chromium, and nickel, having approximately 12.4% chromium. In an assay to discover how much chromium is present, the steel is oxidize in acid to form chromate ion. The resulting solution is treated with excess lead(II) ion to produce a stable precipitate of lead chromate, according to the following reaction:

CrO₄^2-(aq) + Pb²⁺(aq) ==> PbCrO₄(s)

How many mL of 12.2% lead (II) nitrate solution (density=1.39 g/mL) should be used to assure 50% excess Pb2+ for the reaction involving a 3.566 gram stainless steel sample?

I keep getting 24 mL, but the correct answer is 25 mL. please help thanks!!!

Answer 1

Cr in 3.566gms of stainless steel sample = 3.566 x 12.4 / 100 = 0.442 g.

...........................................................or , = 0.442 / 51.996..... = 8.5 x 10^-3 moles

1mole of CrO4^2- ion is equivalent to 1 mole of Cr

therefore moles of CrO4^2- required for reaction = 8.5 x 10^-3 moles

As per reaction this is also equivalent to 8.5 x 10^-3 moles of Pb(NO3)2

but we use 50% excess Pb(NO3)2 ie. = 4.25 x 10-3 moles

so the total moles of Pb(NO3)2 required to be used = 0.01275 moles

wt. of Pb(NO3)2 in 0.0125 moles of Pb(NO3)2 = 0.01275 x 331.19 = 4.228 gms.

Now calculate the density* of 12.2% solution = 0.169 gms /ml

so the volume of solution required = 25.0 ml

* It is understood that the given density of Pb( NO3)2 solution is the density of 100% Pb(NO3)2 solution.

**************************************************************************************************