Answer to Question #184653 in General Chemistry for Lisa

Question

1.  An 600.0-mg sample of chromium ore was dissolved and the chromium oxidized to chromate ion. The solution was treated with 15.00 mL of 0.1750 M AgNO3. The resulting precipitate of AgCr2O4 was removed and discarded. The excess AgNO3 required 12.50 mL of 0.1240 M KSCN for titration. The pertinent ionic reactions are:

Cr2O42-+ 2Ag+→ Ag2CrO4(s)

Ag++ SCN-→ AgSCN (s)

 

 Calculate the % Cr2O3 in the ore.

Answer 1

Molar Mass of ore = 159.69

600 mg = 600g

600/159.69

= 3.76M

Molar Mass of AgNO₃ = 169.87

0.1750 × 169.87

= 29.73 × 12.5

= 371.591

=(371.591/600)100

= 61.93%