Question

Determine the percent yield of each of the nickel (II) compounds

Part A

Step 1:6.107 g (weight of starting material) [Ni(H2O)6]Cl2

Step 7: 5.292 g (weight of product)

Part B

Step 8: 6.115 g (weight of starting material)

Step 11: 10.892 g (Weight of product)

i dont know if this equation goes any further but it is-

[Ni(H2O)6]Cl2(aq)+6NH3(aq)-----> [Ni(NH3)6]Cl2(aq)+6H2O(l)

coordination compound cooled then precipitated by 95% ethanol

[Ni(H2O)6]Cl2(aq)+3en(aq)-----> [Ni(en)3]Cl2*2H20(aq)+4H2O(l)

The compound is precipitated with 95% ethanol solution.

Answer 1

Part-A: Now, 237.69 [Ni (H₂O)] Cl2 + 6 NH₂ → [Ni(NH3), Cl₂ Molecular + 6H2O weight: 237.69 g amount Taken: Tozg a As [Ni(M, 6)1 cz gives [Ni(wHs)]cz , here I mol [Ni(Ho) , ] > gives 1 mol [Ni(NH3) 6] a Bows 1029 [Ni(1,0), ]c, was taken. 51079 [Ni(470),] l, = 107 = 0.450 moli 0.450 mol [Ni(H2O) was taken to react with NH₂ to give 0.450 mo [Ni(NH3) ] 12. . If the whole reaction takes place, the amount of [N: (NH3) ] 2 should be (0.450 x 231-78) g. = 104.34 9 but we get product = 5.292 g (Practical yield) and theoritical yield = 104.34 q . .. wield, :') yields thron Practical yield wody (theoritical yield" E =( 5.292 X 100)% = 5.07 % 104.34 | | Ai la f [N (NH) 144 = 5-07 / / X 100

ni (120),]&2 + 3 en —[Ni(en), 7 cl. 253 en = Nra a Nha 1 + 4630 345.99 g + 4H₂O molecular 237.69 g weight :- 6.1159 amount. Taken As 1 mol of [Ni(en), Cl2, 240 was given by 1 mol of [Ni(H20)]42 Now, 6.115 9 [Ni(H₂O) Cl2 = 6-115 237.69 = 0.0257 mol au 0.257 mol [Ni(H₂O) 67 Cl22 H₂0 gives 0.257 mol 0 [Ni(en)3748.2830 . amount of [Ni(en)z] es. 2H20 = (00257 x 345-99), = 88.90129 If whole amount of [Ni (H₂O) ] C2 converted into product then amount of product obtained = 8.90129 But we get more amount of product (10,8924) than we expect to get (8.90129) & There must be some thing wrong.. Sonry !!