Stoichiometry Lab

Question

• CuCO3^.Cu(OH)₂--> CO₂+H₂O+2CuO

How many grams of Copper II oxide are produced if you start with 1.01 g of Copper II carbonate.

Answer 1

the balanced chemical equation that we're starting with here is

CuCO3​ ⋅ Cu(OH)2​ → CO2 ​ + H2​O + 2CuO

meaning that the mole ratio is 1 mole CuCO3: 2 moles of CuO

next then i would determine the moles of CuCO3 by dividing the given mass by the molar mass

Moles = 1.01 g / 221.114 g/mol = 0.004568 mol CuCO3

using the mole ratio that becomes

Moles of CuO = 0.004568 mol CuCO3 x 2 mol CuO / 1 mol CuCO3 = 0.009136 mol CuO

the last step then is multiplying that value by the molar mass

0.009136 mol CuO x 79.545 g/mol = 0.727 g

meaning that you end up producing approximately 0.727 grams of CuO in this scenario

Answer 2

CuCO3.Cu(OH)2--> CO2+H2O+2CuO

 

Based upon the reaction given, I am assuming that CuCO3 combines with Cu(OH)2 to give us CO2,H2O and CuO.

 

The molecular mass for CuCO3 is 111.54 gms while that of CuO is 79.5 gms.

 

For every 111.54 gms of Copper II carbonate (CuCO3) we producing 2x79.5 gms of CuO.

 

Therefore for 1.01 gms of CuCo3 we should get 1.44 gms of CuO.