A. Chemistry of Gold/Mining. Please provide complete correct and detailed mathem

Question

A. Chemistry of Gold/Mining. Please provide complete correct and detailed mathematical solution done on computer or by hand. This is all the information provided. Please only provide answers. Thank you.

You are a plant manager for Grubstake Mining Company, a company that mines gold ores and extracts the gold from them by cyanide leaching process. The leeching plant is your responsibility. Recently, you have become suspicious of Mr. Johnson, the foreman of the night shift, who has bought a Lear Jet, and his brother, the security guard, who is driving to work in a Ferrari Testarossa. Secretly you gather data for a material balance on the gold. Over a period of one week, 15,291 short tons of ore enter the plant and the average assay is 0.12 troy ounce of gold per ton of ore. During this week there is no change in the inventory of gold bullion and 1500 troy ounces of gold bullion are shipped You measure and analyze the various solids and solutions in the plant throughout the week and find no significant change, except for one large tank containing (throughout the week) 10,000 U.S. gallons of solution. At the start of the week the solution analyzes 0.050 wt % NaAu(CN)2, but at the end of the week this has risen to 0.088 wt %; the density of the solution is 1 g/cm. Are you suspicions justified, or did Johnson & Johnson inherit a fortune from a rich relative in the baby powder business, as they claim? 3

Explanation / Answer

Assay of gold per ton of ore = 0.12 troy ounce

therefore total goid in ore collected in a week = 0.12 x 15291 = 1834.92 troy ounce = 1834.92 x 31.10 g

= 57.07 x 103 g (as troy ounce = 31.10 g)

Now in the tank there is 10,000 US gallon = 10,000 x 3.8 L = 10,000 x 3.8 x 1000 cm3.

Therefore mass of the tank solution is = (10,000 x 3.8 x 1000 cm3 )x 1 g/ cm3 = 3.8 x 107 g

Hence mass of NaAu(CN)2 in tank added during the week = 3.8 x 107 g x (0.088 - 0.050)/ 100 = 14.4 x 103 g

Now molecular wt of NaAu(CN)2 is 272 and atomic mass of Au is 197

Hence amount of gold in the tank = 14.4 x 103 g x (197/272) = 10.43 x 103 g

Given that amount of gold shipped in the week = 1500 troy ounce = 1500 x 31.10 g = 46.65 x 103 g

So the total gold = 10.43 x 103 g + 46.65 x 103 g = 57.08 x 103 g

this is same as the amount of gold expected to be in the ore collected. So suspicion not justified

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