We are asked how much of this rock (in kg) must be processed to obtain 5.1 metric tons of lead.

Calculate mass percent of lead in each:

PbS:

$\mathbf{\%}\mathbf{}\mathbf{Pb}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{MM}\mathbf{\hspace{0.17em}}\mathbf{Pb}}{\mathbf{MM}\mathbf{}\mathbf{PbS}}\mathbf{\times}\mathbf{100}\phantom{\rule{0ex}{0ex}}\mathbf{\%}\mathbf{}\mathbf{Pb}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{207}\mathbf{.}\mathbf{2}}{\mathbf{239}\mathbf{.}\mathbf{27}}\mathbf{\times}\mathbf{100}$

**% Pb = 86.60 %**

PbCO_{3}

$\mathbf{\%}\mathbf{}\mathbf{Pb}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{MM}\mathbf{\hspace{0.17em}}\mathbf{Pb}}{\mathbf{MM}\mathbf{}{\mathbf{PbCO}}_{\mathbf{3}}}\mathbf{\times}\mathbf{100}\phantom{\rule{0ex}{0ex}}\mathbf{\%}\mathbf{}\mathbf{Pb}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{207}\mathbf{.}\mathbf{2}}{\mathbf{267}\mathbf{.}\mathbf{21}}\mathbf{\times}\mathbf{100}$

**% Pb = 77.54 %**

PbSO_{4}

$\mathbf{\%}\mathbf{}\mathbf{Pb}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{MM}\mathbf{\hspace{0.17em}}\mathbf{Pb}}{\mathbf{MM}\mathbf{}{\mathbf{PbSO}}_{\mathbf{4}}}\mathbf{\times}\mathbf{100}\phantom{\rule{0ex}{0ex}}\mathbf{\%}\mathbf{}\mathbf{Pb}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{207}\mathbf{.}\mathbf{2}}{\mathbf{303}\mathbf{.}\mathbf{27}}\mathbf{\times}\mathbf{100}$

**% Pb = 68.32 %**

Calculate mass % for the rock:

$\mathbf{\%}\mathbf{}\mathbf{Pb}\mathbf{}\mathbf{=}\mathbf{}(0.38)(86.60)\mathbf{+}(0.25)(77.54)\mathbf{+}(0.174)(68.32)$

**% Pb = 64.18 %**

Lead is found in Earths crust as several different lead ores. Suppose a certain rock is composed of 38.0% PbS (galena), 25.0% PbCO_{3} (cerussite), and 17.4% PbSO_{4} (anglesite). The remainder of the rock is composed of substances containing no lead.

How much of this rock (in kg) must be processed to obtain 5.1 metric tons of lead? (A metric ton is 1000 kg.)

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