the Cr atoms are not balanced in the 1st equation
...probably the source of your trouble
chromate vs. chromite
Lead(II) chromate (PbCrO4) is used as a yellow pigment to designate traffic lanes, but has been banned from house paint because of the potential for lead poisoning. The compound is produced from chromite (FeCr2O4), an ore of chromium:
FeCr2O4 + K2CO3 + O2 -> Fe2O3 + K2CrO4 + CO2
Lead(II) ion then replaces the K+ ion. If a yellow paint is 0.511% PbCrO4 by mass, how many grams of chromite are needed per kilogram of paint?
-> So replacing the K+ with the Pb2+ ion, the equation becomes:
FeCr2O4 + PbCO3 + O2 -> Fe2O3 + PbCrO4 + CO2
0.511% = (mass of PbCrO4/mass of paint)*100% = (0.511g/100g)*100%
(0.511g/100g)(1000g/1kg) = 5.11g PbCrO4/kg of paint
I calculated the molar mass of FeCrO4 = 171.84g/mol
And the molar mass of PbCrO4 = 323.19g/mol
So then,
(5.11g PbCrO4)(171.84g FeCr2O4/323.19g PbCrO4) = 2.716985055g FeCr2O4 = 2.72g FeCr2O4
...But the answer key says 1.77g FeCr2O4. So the obvious question here is, what could I have done wrong?
1 answer