Mining for Gold Unlike most metals, gold occurs in nature as...

Mining for Gold Unlike most metals, gold occurs in nature as the pure element. Miners in California in 1849 searched for gold nuggets and gold dust in streambeds, where the denser gold could be easily separated from sand and gravel. However, larger deposits of gold are found in veins of rock and can be separated chemically in the following two-step process: (1) $4 \mathrm{Au}(s)+8 \mathrm{NaCN}(a q)+\mathrm{O}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(\ell) \rightarrow$ $$4 \mathrm{NaAu}(\mathrm{CN})_{2}(a q)+4 \mathrm{NaOH}(a q)$$ (2) $2 \mathrm{NaAu}(\mathrm{CN})_{2}(a q)+\mathrm{Zn}(s) \rightarrow$ $$2 \mathrm{Au}(s)+\mathrm{Na}_{2}\left[\mathrm{Zn}(\mathrm{CN})_{4}\right](a q)$$ If 23 kilograms of ore is $0.19 \%$ gold by mass, how much Zn is needed to react with the gold in the ore? Assume that reactions 1 and 2 are $100 \%$ efficient.

Mining for Gold Unlike most metals, gold occurs in nature as the pure element. Miners in California in 1849 searched for gold nuggets and gold dust in streambeds, where the denser gold could be easily separated from sand and gravel. However, larger deposits of gold are found in veins of rock and can be separated chemically in the following two-step process:
(1) $4 \mathrm{Au}(s)+8 \mathrm{NaCN}(a q)+\mathrm{O}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(\ell) \rightarrow$
$$4 \mathrm{NaAu}(\mathrm{CN})_{2}(a q)+4 \mathrm{NaOH}(a q)$$
(2) $2 \mathrm{NaAu}(\mathrm{CN})_{2}(a q)+\mathrm{Zn}(s) \rightarrow$
$$2 \mathrm{Au}(s)+\mathrm{Na}_{2}\left[\mathrm{Zn}(\mathrm{CN})_{4}\right](a q)$$
If 23 kilograms of ore is $0.19 \%$ gold by mass, how much Zn is needed to react with the gold in the ore? Assume that reactions 1 and 2 are $100 \%$ efficient.

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Key Concepts

Reaction Efficiency (Percent Yield)

Assuming 100% reaction efficiency implies that every valid mole of reactant converts completely into product without any losses. This assumption simplifies calculations by eliminating side reactions or incomplete conversions, making it easier to directly apply stoichiometric ratios from the balanced chemical equations.

Stoichiometry

Stoichiometry is the calculation of reactants and products in chemical reactions using the mole relationships dictated by balanced equations. It allows one to determine how much of each reactant is required or produced, making it essential for quantifying substances in any chemical process.

Percent Composition by Mass

Percent composition by mass refers to the percentage of a particular element or compound present in a mixture or ore. It is calculated as the mass of the component divided by the total mass of the sample, and it is crucial for determining the amount of a specific substance available for reaction in a given mass of material.

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