Buchner filtration set-up to filter crystals
A student measures a yield of 1.2 g of Cu(NH3)SO4·H2O after carrying out the method mentioned above
Answers
???1. CuSO4?+ 4NH3?→ Cu(NH3)4SO4?H2O +4H2O
Remember that as we have dissolved the copper sulfate in solution this is actually hydrated copper(II) sulfate CuSO4?5H2O = 249.5
???2. CuSO4?5H2O = 249.5
Cu(NH3)4SO4?H2O?= 245.5
???3. 0.00601 moles (based on 1.5 g of CuSO4?5H2O)
???
4. Theoretical yield of Cu(NH3)4SO4?H2O = 1.48 g
As there is a molar ratio of 1:1 we know we will have 0.00481 moles of Cu(NH3)4SO4?H2O
The theoretical mass that we would expect to produce can then be calculated
???5. Percentage yield =
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