Nick and Kel reacted 2.80 g of NiCl2· 6 H2O with 7.00 mL of 4.00 M C2H8N2. They recovered 2.40 g of product, [Ni (C2H8N2) 3] Cl2. At the end, the filtrate was a dark blue solution. The litmus test on the filtrate solution left the red litmus paper unchanged in color.

Question

Chemistry

Guillermo Huang

3 February, 08:37

Nick and Kel reacted 2.80 g of NiCl2· 6 H2O with 7.00 mL of 4.00 M C2H8N2. They recovered 2.40 g of product, [Ni (C2H8N2) 3] Cl2. At the end, the filtrate was a dark blue solution. The litmus test on the filtrate solution left the red litmus paper unchanged in color.

+1

Answers (1)

Know the Answer?

Miss Piggy · Answer 1

Actual Product yield = 0.028 x 308.7

Explanation:

Limiting reactant

moles of NiCl2.6H2O = 2.80 g/237.7 g/mol = 0.011 mol

moles of C2H8N2 = 4 M x 7.00 ml = 0.028 mol

for formation of complex,

If all of NiCl2.6H2O is consumed we would need = 0.011 x 3 = 0.033 mols of C2H8N2

If all of C2H8N2 is consumed we reuire = 0.028/3 = 0.009 mols ofNiCl2.6H2O

since moles of NiCl2.6H2O available is in excess, C2H8N2 is limiting reactant

Actual Product yield = 0.028 x 308.7