Gold is isolated from rocks by reaction with aqueous cyanide, \(\mathrm{CN} : 4 \mathrm{Au}(s)+8 \mathrm{NaCN}(a q)+\mathrm{O}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow\) \(4 \mathrm{Na}\left[\mathrm{Au}(\mathrm{CN})_{2}\right](a q)+4 \mathrm{NaOH}(a q) .\) (a) \(\mathrm{Which}\) atoms from which compounds are being oxidized, and which atoms from which compounds are being reduced? (b) The \(\left[\mathrm{Au}(\mathrm{CN})_{2}\right]^{-}\) ion can be converted back to \(\mathrm{Au}(0)\) by reaction with \(\mathrm{Zn}(s)\) powder. Write a balanced chemical equation for this reaction. (c) How many liters of a 0.200\(M\) sodium cyanide solution would be needed to react with 40.0 \(\mathrm{kg}\) of rocks that contain 2.00\(\%\) by mass of gold?

In the given reaction, \(4 \mathrm{Au}(s)+8 \mathrm{NaCN}(a q)+\mathrm{O}_{2}(g)+\mathrm{H}_{2}\mathrm{O}(l) \longrightarrow 4 \mathrm{Na}\left[\mathrm{Au}(\mathrm{CN})_{2}\right](a q)+4 \mathrm{NaOH}(a q)\), we must determine which atoms from which compounds are being oxidized, and which atoms from which compounds are being reduced. The oxidation states of the atoms are given as follows: Au: 0 (solid gold) NaCN: Na (+1), C (-3) and N (-2) (the sum is equal to 0) O2: 0 (diatomic oxygen) H2O: H (+1) and O (-2) (the sum is equal to 0) Na[Au(CN)2]: Na (+1), Au (+1), C (-2), N (-1) (the sum is equal to -1) NaOH: Na (+1), O (-2), and H (+1) (the sum is equal to 0) Comparing the oxidation states in the reactants and products, we see that the Au atom goes from oxidation state 0 to +1, and the O atom goes from oxidation state 0 to -2. This means that Au is being oxidized, and O2 is being reduced. (a) The Au atom in solid gold is being oxidized, and the O atom in molecular oxygen is being reduced.

Now, we will write a balanced chemical equation for the conversion of the gold cyanide complex back to gold by reacting with zinc powder. \(\mathrm{Zn}(s)+2\left[\mathrm{Au}(\mathrm{CN})_{2}\right]^{-}(a q) \longrightarrow 2 \mathrm{Au}(s)+\left[\mathrm{Zn}(\mathrm{CN})_{4}\right]^{2-}(a q)\) This is a balanced equation for the reaction. (b) The balanced chemical equation for the reaction of the gold cyanide complex with zinc powder is \(\mathrm{Zn}(s)+2\left[\mathrm{Au}(\mathrm{CN})_{2}\right]^{-}(a q) \longrightarrow 2 \mathrm{Au}(s)+\left[\mathrm{Zn}(\mathrm{CN})_{4}\right]^{2-}(a q)\).

We are given that the rocks have 2.00% by mass of gold and that there are 40.0 kg of these rocks. To calculate the volume of a 0.200 M sodium cyanide solution needed to react with this gold, we first need to find the moles of gold in the rocks: mass of gold in rocks = 40.0 kg * (2.00 / 100) = 0.800 kg Now, we convert the mass of gold to moles using the molar mass of Au (197 g/mol): moles of gold = (0.800 kg * 1000 g/kg) / 197 g/mol ≈ 4.065 moles Using the stoichiometry of the given reaction, we find that 8 moles of NaCN are needed for every mole of gold: moles of NaCN needed = 4.065 moles Au * (8 moles NaCN / 1 mole Au) ≈ 32.520 moles NaCN With a concentration of 0.200 M NaCN solution, we can find the volume of solution needed: volume of NaCN solution = moles of NaCN / concentration volume of NaCN solution = 32.520 / 0.200 ≈ 162.6 L (c) The volume of the 0.200 M sodium cyanide solution needed to react with 40.0 kg of rocks containing 2.00% by mass of gold is approximately 162.6 liters.