LAB 8: CU ISOLATION Students can use K, to determine product losses in washing s
ID: 550753 • Letter: L
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LAB 8: CU ISOLATION Students can use K, to determine product losses in washing steps. Students use percent yield to determine efficiency of lab procedure. OBJECTIVES s to determine the · Students can add equations that describe multple reaction thermodynamic parameters of the combined reactions (the final equation). which a * Students can use t-asmA T, and G to determine temperature at reaction is spontaneous To the left is a picture of the Berkley Mine Pit near Butte, Montana. Formerly an open pit copper mine, the mine shut down operations in 1982 and since then ground water has filled the mine. Over time the dissolution of transition metal ions has led to the creation of an acidic pool of water that is laden with Cu", Cd, and Zn ions. As the metal ion content builds in the water the pH of the water lowers which increases the capacity of the water to oxidize more metals making them water soluble and thus, the water continues to lower in pH. Pumps that removed water from the pit have been INTRODUCTION Photo of Berkley Mine Pit turned off so, it's also estimated that the water will rise until it reaches the natural water table (estimated for 2020). When this occurs, the pit will reverse flow and start to drain back into the ground water contaminating the water around the city of Butte. Thus, Berkeley has been designated an EPA Superfund site and plans have been formed to remediate the site. One strategy is to recover the copper from the pit and sell it to help finance further clean-up efforts. Currently there are approximately 1.7 x 10 liters of water in the pit with copper concentrations of 180 ppm Cu. With 2015 copper prices at $00.05/g, the recovery and sale of copper technicians will apply a methodology to selectively precipitate copper ions from might not only finance the cleanup it might even turn a profit. In this lab, chemical water and convert them with thermal reduction to pure copper During part 1 of the lab, technicians will precipitate basic copper carbonate, Cu2(OH)2CO,(s) out of a 1.0M solution of copper II sulfate (Equation 1 and 2). This precipitate is also the predominate compound in the mineral malachite or mal- achite green. In the second part of the lab, technicians will heat the basic copper carbon will turn to carbon dioxide and the copper II will be converted to copper overall yield. carbonate which will force a conversion to copper (II) oxide (Equation 3). Finally, a redox reaction will be performed in the presence of carbon powder such that the metal (Equation 6). Students will evaluate efficiency of each step and determine Lab 8: Cu Isolation 65Explanation / Answer
1. The first thing to do is to remember the formula of parts per million
ppm = miligrams of solute / kg of solvent
We have liters of water and we need kilograms, since nothing else is provided I will assume that the density of water is 1 Kg / L (pure water) so we have 1.7 x 109 kg of water
ppm = 180 = mg of Cu / 1.7 x 109 kg of water
mg of Cu = 180 * 1.7 x 109 = 3.06 x 1011 mg of Cu
g of Cu = 3.06 x 108 grams
price = 3.06 x 108 * 0.05 = $ 1.53 x 107 dollars
2. According to your reaction you need 2 moles of sodium carbonate to react 2 mole of sulphate carbonate to produce 1 mole of copper carbonate
The precipitate of copper carbonate has a molecular weight of 221 g/gmol
moles of copper carbonate is = mass / molecular weight = 2 / 221 = 0.009 moles we want
we need 2 moles of every reactant for every mole of precipitate
0.009 * 2 = 0.018 moles we want
Molarity of reactants = 1M
Molarity = moles / Volume
1 M = 0.018 / V
V = 0.018 / 1 = 0.018 L = 18 ml of each reactant to produce 2 grams of precipitate
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