Question

Determine the percent yield of copper.

3. What is the theoretical yield of copper produced by this reaction? Materials: Copper (II) chloride • Aluminum foil • Stirring rod • Two 100 ml beakers • Graduated cylinder • Safety glasses • Spatula or scoopula • Balance Filter paper front counter Funnel • Watch glass Ring stand • Iron ring Safety: Safety glasses must be worn at all times! Copper (II) chloride is an irritant and moderately toxic. Avoid inhaling the powder and ingestion. • Wash hands thoroughly after completing the lab. → The reaction is exothermic and produces a lot of heat. Follow procedure carefully. • DO NOT look into the beaker directly. Observe the reaction from the side. co to Procedure: to hundrets place 1. Weigh approximately 3.5 g of CuCl2 and place them in the 100 ml beaker. Record the exact mass in the data table below. Note the appearance of the CuCl2 crystals in the observation section. FIS KEY 2. Add 50 mL of distilled water to the beaker containing the CuCl2 crystals. Stir the solution until all of the CuCl2 crystals have dissolved. Record any Want enne vowme observations in the observation section. Record the final volume of this in grauak ylinder forecord solution. toward ths place chhure Volumebt 3. Weigh out approximately 0.2 g of aluminum foil. Record the exact mass in when the data table. 4. Loosely crumple the piece of aluminum foil into a ball. Carefully place it into the CuCl2 solution. Stir the reaction occasionally. Record your observations of the resulting reaction. Set for at least 20 minutes 5. Set up a ring stand with an iron ring securely attached to it. Place a funnel in the iron ring. Cucia.at could effect which is 3.59 reagent Als) +

Answer 1

In this experiment, the balanced chemical reaction that takes place is-

3 CuCl₂ + 2 Al    ----->   2 AlCl₃ + 3 Cu

Mass of CuCl₂ taken = 3.50 g

molar mass of CuCl₂ = 134.45 g/mol

Mass of Al = 0.20 g

molar mass of Al = 26.98 g/mol

Mass of Cu obtained from 3.50 g of CuCl₂ is

$3.5gCuCl_{2}\left ( \frac{1molCuCl_{2}}{134.45gCuCl_{2}} \right )\left ( \frac{3molCu}{3molCuCl_{2}} \right )\left ( \frac{63.54gCu}{1molCu} \right )=1.65 gCu$

But actual amount of Cu obtained = (mass of filter paper + solid) - (mass of filter paper) = 2.74 - 1.11 = 1.63 g

Therefore, percent yield =(actual yield/ theoretical yield)x 100

$%$$yield = \frac{1.63}{1.65} \times 100=98.78$

Therefore, percent yield = 98.78%

(Note: amount of Al weighed seems to be very low with just 0.20 g of Al foil, only .706 g of Cu must be obtained.)