The gold in the zinc replacement solution is carried out according to the following reaction formula: 2Au(CN) 2 - +Zn=2Au↓+Zn(CN) 4 2 - The reaction is rapid and the replacement is complete. When the cyanide concentration and the alkali concentration in the solution are small, the oxygen dissolved in the solution redissolves the precipitated gold and oxidizes the zinc to form a hydroxide precipitate. The Na 2 Zn(CN) 4 formed in the above reaction also decomposes to form a zinc cyanide precipitate: Zn+ Na 2 Zn(CN) 4 +Zn(OH) 2 =2Zn(CN) 2 ↓+2NaOH The resulting zinc hydroxide and zinc cyanide form a white film precipitate on the surface of the metal zinc, which prevents the gold and silver from completely precipitating from the cyanide solution. In solutions with higher cyanide and base, in addition to the formation of Zn(CN) 4 2 - complex anions, zinc also releases hydrogen from the well as follows: 4NaCN+Zn+2H 2 O=Na 2 Zn(CN) 4 +2NaOH+H 2 ↑ 2NaOH+Zn=Na 2 ZnO 2 +H 2 ↑ This reaction increases the consumption of zinc and releases a large amount of oxygen. However, hydrogen reacts with dissolved oxygen in the solution to form water, which can reduce or even prevent the dissolution of the precipitated gold. Metal zinc can also be oxidized. Under normal zinc powder replacement conditions, the concentration of cyanide should be controlled to about 0.02% and about 0.01% of calcium oxide in the gold-containing solution entering the displacement sedimentation tank. When zinc wire is replaced, cyanide and alkali concentrations are correspondingly higher because some plants do not degas the solution. Of course, it is preferable to remove the dissolved oxygen in the solution through the degassing tower before the zinc-containing solution is replaced by the gold-containing solution to completely eliminate the harmful influence on the gold of the replacement precipitate. Typically the cyanide solution containing less amount of lead, since lead can be improved in conjunction with zinc precipitation of gold, it is often added to an appropriate amount of lead acetate or lead nitrate to the mother liquor. However, excessive lead will cause zinc to be consumed due to many edge reactions, and the precipitation of gold will be slow and incomplete, or the precipitate will be contaminated by the formation of Pb(OH) 2 precipitate, so it is only for each ton of mother liquor. Add 5 to 10 g of lead nitrate. The presence of copper produces a metallic copper precipitate that consumes zinc. Mercury will form an alloy with zinc. The presence of sulfur ions produces ZnS and PbS precipitates that contaminate the metal zinc. Since the cyanide solution contains calcium and hydroxide ions, the presence of nickel can seriously affect the precipitate. Therefore, the nickel in the Kurr Addison plant's lean liquid is nearly 90 × 10 -6 . The relationship between zinc precipitation method and sodium cyanide concentration, oxygen concentration and gold recovery rate shows that when the cyanide solution is gold 15mg/L, NaCN 0.015%-0.07%, NaOH 0.015%, oxygen 0-3.1mg ∕L, the amount of zinc added is 1g ∕L. When the NaCN concentration is increased, the consumption of zinc is increased due to the tendency to form a precipitate. When the solution contains 1 mg ∕L of oxygen, the recovery rate of gold can reach 97%-100%, and when the oxygen content increases to 30 mg ∕L, the recovery rate of gold is only 78%-80%. The mustard packaging tube is a popular choice for many manufacturers and consumers because of its convenience and ease of use. The tube is easy to squeeze, allowing users to dispense the mustard easily and accurately. It is also lightweight, making it easy to carry and transport.
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O 2 +H 2 O=Zn(OH) 2 ↓
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