Question

Weight of evaporated dish= 48.32 g    Weight of evaporating dish= 36.89 g

Weight of evaporating dish and watch glass and NaHCO3(s)=49.48g Weight of evaporating dish and CuCO3*C(OH)2=39.33 g   

Actual weight of NaHCO3(s) reacted= 1.75 g    Actual weight of CuCO3*Cu(OH)2 reacted= 2.44 g

Weight of evaporating dish and watch glass and NaCl(s)=49.11 g    Weight of evaporating dish and watch glass and CuO(s)=38.83 g

Actual weight of NaCl(s) produced=0.88g    Actual weight of CuO(s) produced= 1.94 g

Sodium Carbonate with Hydrochloric Acid.    Basic Copper (2) Carbonate.

So far these are the results of my Chemistry lab, I am having trouble trying to find the Actual Moles produced and reacted for each the Sodium Carbonate with Hydrochloric Acid and Basic Copper (2) Carbonate. As well as the Theoretical weight and moles of NaCl(s) produced and the Theoretical weight and moles of CuCO(s). I also need help with balancing out the equations for the two reactants CuCO3*Cu(OH)2.

Answer 1

The balanced reaction is

CuCO3+HCl+NaHCO3= CuO+CO2+NaCl+H2CO3

which comes by adding the reactions CuCO3+2HCl=CuCl2+H2CO3 and NaHCO3+HCl+NaCl+H2CO3

Moles of NaCl produced = 0.88/58.44 = 0.015 .

Moles of CuO produced = 1.94/79.545 = 0.0243

Moles of Cu carbonate reacted = 2.44/ molecular wt of CuCO3 ( here there is a confusion . If one takes the molecular wt of the carbonate long with its hydrate , then the no of moles will be different. I am taking CuCO3 leaving out the CuOH2 part) = 2.44/221.116 moles= 0.011 moles.

Moles of Bicarbonate reacted = 1.75/84.007 = 0.020 moles.

According to the reaction one mole of CuCO3 gives on emole of CuO, thus 0.011 moles of CuCO3 should have given 0.011 mole of CuO. But the true moles of CuO produced is 0.0243. Here o.011 mole is the theoretical yeild of the CuO and the yeild truely obtained is the true yeild which is 0.0243 moles of CuO here.