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If the student collects 234.6 g of product, what is the percent yield of [Cu(NH3)4H2O]SO4?

If the student collects 234.6 g of product, what is the percent yield of [Cu(NH3)4H2O]SO4?

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Answer #1

The possible reaction will be

Assuming that one mole of CuSO4.5H2O and 4 moles of NH3 is present

Molar mass of [Cu(NH3)4H2O]SO4 = 245.74 gm/mol

Percent Yield of Reaction = Actual Mass/Theoritical Mass * 100

=> 234.6/245.74 * 100

=> 95.46%

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