lead sulfide, in ores, can be assayed by the reaction:
2PbS + 6HNO3 (aq) + K2Cr2O7(aq)--> 2 S + 2PbCrO4(s)+ 4NO2(g) + 2KNO3(aq) + 3H2O
A 6.053 g sample of ore, treated with excess nitric acid and potassium dichromate yielded 6.094 g of lead (II) chromate. Calculate the percent by mass of lead sulfate in the ore.
if you know how to do this...can you show me your work. i did it and i ended up with 37.5 %. I don't know if I did this right...I think I did something wrong with the mole ratio. Thank you so much =]
2007-10-27 14:11:39 · 2 answers · asked by Save A Tree [Remove a Bush] 4 in Science & Mathematics ➔ Chemistry
PbS is 239.266 g/mol; PbCrO₄ is 323.1937 g/mol; 6.094 g / 323.1937 g/mol PbCrO₄ = 18.85 mmol PbCrO₄; 18.85 mmol PbS x 239.266 mg/mmol PbS = 4510 mg = 4.510 g PbS
4.510 x 100 / 6.053 = 74.51% to the correct 4 significant figures
2007-10-27 14:30:46 · answer #1 · answered by Anonymous · 1⤊ 0⤋
Atomic weights: Pb=207 S=32 PbS=239 Cr=52 O=16 PbCrO4=323
Let PbCrO4 be called LC
6.094gLC/6.053gOre x 1molLC/323gLC x 2molPbS/2molLC x 239gPbS/1molPbS x 100% = 74.5%
2007-10-27 21:22:37 · answer #2 · answered by steve_geo1 7 · 0⤊ 1⤋