Asked by Anonymous
A stainless steel alloy is to be analyzed for its chromium content. A 5.50 g sample of the steel is used to produce 250.0 mL of a solution containing Cr2O72−. A 10.0-mL portion of this solution is added to BaCl2(aq). When the pH of the solution is properly adjusted, 0.145 g BaCrO4(s) precipitates.
Answers
Answered by
DrBob222
What's the question. I'm assuming you want to calculate the percent Cr but not sure of that.
steel --> 2Cr^3+ --> [Cr2O7]^2- --> 2CrO4^- ==> 2BaCrO4
5.5 g.......in 250 cc..........10 mL............................... 0.145 g
Convert 0.145 g BaCrO4 to g Cr^3+. 0.145 g BaCrO4 x (mm Cr/mm BaCrO4) = mm = molar mass.
0.145 x (52/253.3) = 0.0298 in the 10 mL sample. The amount in the 250 cc flask is
0.0298 x (250/10) = 0.744 g in the 5.5 g sample.
% Cr = 0.744/5.5)*100 = ?
steel --> 2Cr^3+ --> [Cr2O7]^2- --> 2CrO4^- ==> 2BaCrO4
5.5 g.......in 250 cc..........10 mL............................... 0.145 g
Convert 0.145 g BaCrO4 to g Cr^3+. 0.145 g BaCrO4 x (mm Cr/mm BaCrO4) = mm = molar mass.
0.145 x (52/253.3) = 0.0298 in the 10 mL sample. The amount in the 250 cc flask is
0.0298 x (250/10) = 0.744 g in the 5.5 g sample.
% Cr = 0.744/5.5)*100 = ?
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