Question

Exp 7. Preparation of Copper (I) Chloride Name ________________________ Experiment replacement assignment. The process of preparing copper (I) chloride, CuCl, from metallic copper consists of the reactions listed below:

Rxn 1. Cu (s) + 4H+ (aq) + 2NO3 – (aq) --> Cu2+ (aq) + 2NO2 (g) + 2H2O(l)

Rxn 2. omitted, as it does not involve copper

Rxn 3. Cu2+ (aq) + CO3 2- (aq) ---> CuCO3 (s)

Rxn.4 CuCO3 (s) + 2H+ (aq) + 4Cl– (aq) ---> CuCl4 2- (aq) + CO2 (g) + H2O (l)

Rxn 5. CuCl4 2- (aq) + Cu (s) + 4Cl– (aq) ---> 2 CuCl4 3- (aq) additional copper

Rxn 6. CuCl4 3- (aq) --> CuCl (s) + 3Cl - (aq)

1. Analyze all reactions and “follow” the initial cupper as it changes into different forms until the final product, CuCl, is made. List all of these forms, in the proper order; include the physical states.

2. Metallic copper used in Rxn1 is the limiting reactant in this process. All other reactants in all steps are available in sufficient amounts. If 2.00 g of copper is used as the starting material, how many grams of the final product, CuCl, can be made?

3. If the yield of this overall process is 75.0%, what mass, in grams, of CuCl amount is produced when from 2.00 g of the initial Cu?

4. Nitrogen dioxide is the toxic product of Rxn 1. Calculate what volume, in L, of NO2 is produced by complete reaction of 2.00g of copper if the gas is collected at the pressure of 745.0 mm Hg and temperature of 22.0 o C.

5. Another way to write a balanced equation for reaction 1 is: Cu (s) + 4 HNO3 (aq) --> Cu(NO3)2 (aq) + 2 NO2 (g) + 2 H2O (l) If 55.0 mL of the 3.0 M nitric acid, HNO3, and 2.00 g of Cu were used, which reactant remained after the reaction was complete?

Answer 1

the stages include Cu(s)------>Cu+2(aq), Cu+2(aq)------>CuCO3-(s), CuCO3-(s)-------->CuCl4-2(aq)

CuCl4-2(aq) --------->CuCl(s)

atomic weight of Copper= 63.5 g/mole, moles of Cu= mass/molar mass= 2/63.5 =0.0315

rxn 1. Cu (s) + 4H+ (aq) + 2NO3 – (aq) --> Cu2+ (aq) + 2NO2 (g) + 2H2O(l)

1 mole of Cu gives 1mole of Cu+2. Hence 0.0315 moles of Cu gives 0.0315 moles of Cu+2

from Rxn 3. Cu2+ (aq) + CO3 2- (aq) ---> CuCO3 (s)

1 mole of Cu+2 gives 1 mole of CuCO3, 0.0315 moles of Cu gives 0.0315 moles of CuCo3.

From Rxn-4

Rxn.4 CuCO3 (s) + 2H+ (aq) + 4Cl– (aq) ---> CuCl4 2- (aq) + CO2 (g) + H2O (l)

0.0315 moles of CuCO3 gives 1 mole of CuCl4-2, 0.0315 moles of CuCO3 gives 0,0315 moles of CuCl4-2

from Rxn-5

Rxn 5. CuCl4 2- (aq) + Cu (s) + 4Cl– (aq) ---> 2 CuCl4 3- (aq) additional copper

1 mole of CuCl4-2 gives 2 moles of CuCl4-3, 0,0315 moles if CuCl4-2 gives 2*0.0315= 0.063 moles of CuCl4-3.

fom Rxn-6,

Rxn 6. CuCl4 3- (aq) --> CuCl (s) + 3Cl - (aq)

1 mole of CuCl4-3 gives 1 mole of CuCl

0.063 moles of CuCl gives 0.063 moles of CuCl.

molar mass of CuCl = 99 g/mole, mass of CuCl=0.063*99=6.237 gm . This is the theoretical yield.

when 75% is the actual yield, mass of CuCl produced =6.237*0.75= 4.67 gm

from rxn-1 0.0315 moles of Cu gives 2*0.0315=0.063 mole of NO2

hence in PV= nRT, n= 0.063, V= nRT/P, R =0.0821 L.atm/mole.K, P= Pressure in atm= 745/760 atm =0.98 atm

T= 22 deg.c =22+273= 295K, V= 0.063*0.0821*295/0.98 =1.55 L= 1550ml

3.

moles of HNO3= Molarity* volume in L= 3*55/1000 =0.165 moles, moles of Cu= 0.0315

actual molar ratio of Cu:HNO3 = 0.0315 :0.165 = 1:5.24

Theoretical ratio of Cu: HNO3= 1:4 from reaction is Cu (s) + 4 HNO3 (aq) --> Cu(NO3)2 (aq) + 2 NO2 (g)

hence excess is HNO3. Hence HNO3 remains after the reaction.