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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