Chapter 3: Problem 141
Lead(II) chromate \(\left(\mathrm{PbCrO}_{4}\right)\) is used as the yellow pigment for marking traffic lanes but is banned from house paint because of the risk of lead poisoning. It is produced from chromite \(\left(\mathrm{FeCr}_{2} \mathrm{O}_{4}\right),\) an ore of chromium: $$ \begin{array}{r} 4 \mathrm{FeCr}_{2} \mathrm{O}_{4}(s)+8 \mathrm{~K}_{2} \mathrm{CO}_{3}(a q)+7 \mathrm{O}_{2}(g) \longrightarrow \\ 2 \mathrm{Fe}_{2} \mathrm{O}_{3}(s)+8 \mathrm{~K}_{2} \mathrm{CrO}_{4}(a q)+8 \mathrm{CO}_{2}(g) \end{array} $$ Lead(II) ion then replaces the \(\mathrm{K}^{+}\) ion. If a yellow paint is to have \(0.511 \% \mathrm{PbCrO}_{4}\) by mass, how many grams of chromite are needed per kilogram of paint?
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