Q1: Part A: Determine the theoretical yield of your product in Part A, Cu(NH4)2(

Question

Q1:

Part A: Determine the theoretical yield of your product in Part A, Cu(NH4)2(SO4)2 X 6 H2O (s).

Part B:Determine the theoretical yield ofyour project in Part B, CU(NH3)4SO4 X H2O (s).

Data sheet is attatched. Please show all wokrings. Thanks!!

Experiment 9 - Copper Complexes Winter 2017-201 oso Laboratory Table 3 - All masses in grams (1 mark) Mass of CuSO4 5H20 Co.O2 3.45 Mass of (NH)2SO4 Mass of vial + product Mass of empty vial Mass of Cu(NH)2(SO4)2.6 H2O 21.4S 10.89 Table 4 - All masses in grams (1 mark) 2.53 4.20 Mass of CuSO4 5H20 Volume of NHs (mL) Mass of vial + product Mass of empty vial Mass of Cu(NH3)4S04 H20 13. 7 O. 83 2.9I Sample Vial Label Date Slot #

Explanation / Answer

A)

Number of moles = mass / molar mass

Number of moles of CuSO4.5H2O = 6.02g / 249.685 g/mol

= 0.0241 mol CuSO4.5H2O

Number of moles of (NH4)2SO4 = 3.45 g / 132.14 g/mol

= 0.0261 mol (NH4)2SO4

here moles having small number is the limiting reactant.

molar mass of Cu(NH4)2(SO4)2.6H2O = 399.818 g/mol

Theoretical yield = moles of the limiting reactant x molar mass of product

= 0.0241 mol x 399.818 g/mol

= 9.636 g

Theoretical yield of Cu(NH4)2(SO4)2.6H2O is  9.636 g

B)

Number of moles of CuSO4.5H2O = 2.53 g / 249.685 g/mol

= 0.01 mol CuSO4.5H2O

Volume of NH3 = 4.20 mL = 0.0042 litre

Number of moles of NH3 = Concentration x volume

= 18 mol L1 x 0.0042 litres

= 0.0756 mol  NH3

here moles having small number is the limiting reactant.

molar mass of Cu(NH3)4SO4 .H2O = 245.79 g/mol

Theoretical yield = moles of the limiting reactant x molar mass of product

= 0.01 mol x 245.79 g/mol

= 2.4579 g

Theoretical yield of Cu(NH3)4SO4 .H2O is 2.4579 g

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