Question

Ethyl, Ann & Ami reacted 2.00 g of NiCl₂<span &quot;ms="" reference="" sans="" serif&quot;,="" sans-serif;="" font-size:="" 12pt;"="">·6 H₂O with 7.00 mL of 4.00 M C₂H₈N₂. They recovered 1.68 g of product, (Ni(C₂H₈N₂)₃)Cl₂. What is the theoretical yield? Percent yield?

Answer 1

Molar mass of NiCl₂·6 H₂O = 237.6911 g/mol

Mass of NiCl₂·6 H₂O = 2 g

Moles of NiCl₂·6 H₂O = Mass/Molar mass = 0.00841 moles

Moles of C₂H₈N₂ = Concentration x Volume = 4 x 0.007 L = 0.028 moles

NiCl₂·6 H₂O + 3C₂H₈N₂   = (Ni(C₂H₈N₂)₃)Cl₂

Required moles of C₂H₈N₂ for complete reaction = 3 x 0.00841 = 0.025 moles

Therefore, 0.025 moles of C₂H₈N₂ react with 0.00841 moles of NiCl₂·6 H₂O to produce 0.00841 moles moles of (Ni(C₂H₈N₂)₃)Cl₂

Molar mass of Ni(C₂H₈N₂)₃Cl₂ is 309.8944 g/mol

Therefore, theoretical yield of Ni(C₂H₈N₂)₃Cl₂ = 0.00841 moles x 309.8944 g/mol = 2.61 g

Percent yield= (1.68/2.61) x 100% = 64.4 %