What is the percentage yield if 110.6 g of malachite are formed when 1.5 moles of copper sulfate reacts with an excess of sodium carbonate?Given:- Theoretically, 0.75 moles or 165.84 g of malachite are produced with 1.5 moles of copper sulfateAnd:Measured Mass of CuSO4.5H2O 12.249 g Molar mass of CuSO4.5H2O 249.685 g/mol Moles CuSO4 .5H2O dissolved in reaction mixture(Entry 1 divided by Entry 2) 0.049057813 molMoles of CuSO4 reacted(Same as Entry 3) 0.049057813 molMole to mole ratio of Cu2(OH)2CO3 producedto CuSO4 reacted from balanced chemicalequation (see above) 1 : 2 mol Theoretical moles Cu2(OH)2CO3 produced (Entry 4 times Entry 5) 0.024528907 molMolar mass of Cu2(OH)2CO3 221.11558 g/molTheoretical mass Cu2(OH)2CO3 produced (Entry 6 times Entry 7) 5.423723388 g

Percentage yield is defined as the percent ratio of actual yield. It is calculated to be the actual…

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Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

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What is the percentage yield if 110.6 g of malachite are formed when 1.5 moles of copper sulfate reacts with an excess of sodium carbonate?

Given:

- Theoretically, 0.75 moles or 165.84 g of malachite are produced with 1.5 moles of copper sulfate

And:

Measured Mass of CuSO₄^.5H₂O 12.249 g
Molar mass of CuSO₄^.5H₂O 249.685 g/mol
Moles CuSO₄^.5H₂O dissolved in reaction mixture
(Entry 1 divided by Entry 2) 0.049057813 mol
Moles of CuSO₄ reacted
(Same as Entry 3) 0.049057813 mol
Mole to mole ratio of Cu₂(OH)₂CO₃ produced
to CuSO₄ reacted from balanced chemical
equation (see above) 1 : 2 mol
Theoretical moles Cu₂(OH)₂CO₃ produced
(Entry 4 times Entry 5) 0.024528907 mol
Molar mass of Cu₂(OH)₂CO₃ 221.11558 g/mol
Theoretical mass Cu₂(OH)₂CO₃ produced(Entry 6 times Entry 7) 5.423723388 g

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