Chem 3A -Burdge Worksheet - Chapter 17- Redox Reactions Page 5 of 7 Dissolving a

Question

Chem 3A -Burdge Worksheet - Chapter 17- Redox Reactions Page 5 of 7 Dissolving a Nobel Prize A True Story: During World War II, the Nazis declared no gold should leave Germany, but two Nobel laureates sent their medals to Bohr's Institute in Denmark for protection - a capital offense. Then the Nazis took control of Copenhagen - the capital of Denmark. A chemist, Georgy de Hevesy, decided to dissolve the medals before the Nazis raided the institute. Gold is a very unreactive element that doesn't tarnish or dissolve in much of anything- except aqua regia. When the Nazis ransacked Bohr's Institute, they scoured the building but left the flasks of aqua regia untouched. In 1950, the gold was precipitated from solution and re-cast into medals. Questions 1 and 2 relate to this scenario, 2. Aqua regia is a mixture of nitric acid and hydrochloric acid. Neither component alone can dissolve gold Au), but together they can, according to the balanced reaction below: Au(s) + 4 HCI(aq) + 3 HNOg(aq)- Aucu-(aq) + 3 NO2(g) + H+ (aq) + 3 H2O(aq) (a) (10 pts) A 375.0 mL quantity of concentrated HCI (11.65 M) was mixed with 350.0 mL of concen- trated HNO (15.80 M). What is the maximum amount of solid gold (Au) that could be dissolved in this mixture of acids? Report your answer in grams (b) (3 pts) A 192 g Nobel prize medal made of 24 carat gold (24 carat = 99.95% Au by mass) was added to the above acid solution. Which reagent is the limiting reagent in this scenario? 3. Given the balanced equation in Question 2: (a) (3 pts) What is the oxidation state of N in HNOs? (b) (2 pts) What is the coidizing agent? (c) (2 pts) What is the reducing agent? (d) (3 pts) What element is being reduced?

Explanation / Answer

V = 375 mL of 11.6 M HCl with 15.8 M of V 350 mL HNO3 is added

mol of HCl = MV = 11.65*375 = 4368.75 mmol

mol of HNO3 = M = 15.8 *350 = 5530

ratio is 3:4 so there is excess HNO3

HCl is limiting

4 mmol of HCl acid --> 1 mmol of gold

4368.75_--> 4368.75/4 = 1092.187 mmol of gold = 1.092 mol of gold

mass = mmol*MW = 1.092 *  196.96655 = 215.08 g of Gold

b)

if we added 192 g of medal --> 0.9995*192 = 191.904 g of Gold

then, Gold will limit reaction, according to previous data, we could react up to 215 g of gold

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