Question

Lead(II) chromate (PbCrO₄) is used as the yellow pigment in traffic lanes, but is banned from house paint because of the risk of lead poisoning. It is produced from chromite (FeCr₂O₄), an ore of chromium:

$\mathbf{4}\mathit{F}\mathit{e}\mathit{C}{\mathit{r}}_{\mathbf{2}}{\mathit{O}}_{\mathbf{4}}\mathbf{\left(}\mathit{S}\mathbf{\right)}\mathbf{+}\mathbf{8}{\mathit{K}}_{\mathbf{2}}\mathit{C}{\mathit{O}}_{\mathbf{3}}\mathbf{\left(}\mathit{a}\mathit{q}\mathbf{\right)}\mathbf{+}\mathbf{7}{\mathit{O}}_{\mathbf{2}}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{\to }\mathbf{2}\mathit{F}{\mathit{e}}_{\mathbf{2}}{\mathit{O}}_{\mathbf{3}}\mathbf{\left(}\mathit{s}\mathbf{\right)}\mathbf{+}\mathbf{8}{\mathit{K}}_{\mathbf{2}}\mathit{C}\mathit{r}{\mathit{O}}_{\mathbf{4}}\mathbf{\left(}\mathit{a}\mathit{q}\mathbf{\right)}\mathbf{+}\mathbf{8}\mathit{C}{\mathit{O}}_{\mathbf{2}}\mathbf{\left(}\mathit{g}\mathbf{\right)}$

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?

Short answer

There are 1.77 g FeCr₂O₄of chromite are need to replace per kg of paint.

Step-by-step solution

Calculating the moles of PbCrO4

The moles can be denoted as,

$$\begin{gathered} MolesofPbCr{O}_4=5.11gPbCr{O}_4\times \frac{1molPbCr{O}_4}{323.19gPbCr{O}_4}=0.0158mol \\ 0.0158molPbCr{O}_4\times \frac{1molCr}{1molPbCr{O}_4}\times \frac{1molFeC{r}_2{O}_4}{2molCr}=7.9\times {10}^{-3}mol \end{gathered}$$

Calculating the mass of FeCr2O4

The mass can be calculated as,

$$\begin{gathered} MassofFeC{r}_2{O}_4=7.9\times {10}^{-3}molFeC{r}_2{O}_4\times \frac{223.85gFeC{r}_2{O}_4}{1molFeC{r}_2{O}_4} \\ MassofFeC{r}_2{O}_4=1.77g \end{gathered}$$