Experimental yield lab

Question

CuCO₃ . Cu(OH)₂ --> CO₂+H₂O + 2CuO

How many grams of Copper II oxide are produced if you start wit 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

Assuming the reaction runs to completion and CuCO3 is limiting:

 

1.01g of CuCO3 = 1.01g/(123.546g/mol) = 8.175 mmol

 

So the product moles are 2*8.175 mmol = 16.35mmol

 

From here, we can determine the mass by using molecular weight:

0.01635mol*(79.546g/mol) = 1.3 grams of CuO