Question

POSSIBLE POINTS: In part A, the synthesis of hexamminecobalt(III) chloride, [CO(NH3)6]Cly, your reactants included the following. Notice that these are the values given in the manual, and you can assume you measured it out to the same precision given. • 7.5 mL of 1.5M cobalt(II) chloride solution • 0.3 g of solid ammonium chloride • 0.15 g of activated charcoal (a catalyst) • 7.5 mL of 15 M NH3 . and 7.5 mL of 10% hydrogen peroxide. After synthesizing the cobalt product, you weighed it on a balance and the mass was 2.8245 g. This is the actual yield. What is the percent yield of your cobalt compound? Show your work to obtain full credit. Give simple word equations and then fill in with values, using appropriate units. Bold your final answer and give it with appropriate significant figures. Tips: According to the manual on page 114, the metal is the limiting reagent. You need to calculate the theoretical yield (the maximum amount obtainable of product). You will need to consider the molar relationship of the limiting reagent and product in the reaction on page 110. Then, to determine the percent yield, use the equation actual yield theoretical vid 100.

Answer 1

part A. Sol Given that 7.5mL of 1.5M Cobalt (1) chleride solutim, a39 q solid NaCl, 0.159 q activated chaleoal 7.5mL q isMNH₂ Ý 7.5me of lov. H2O2 were added to Synthesise [Co CNHa)] cl2.(M.wt) = 267.48 g/mol) The following reaction takes place Co, -6 -6 6 + SNH + N 4d >[cow Ha t6 H2O + Ht In words: Cobalt (11) chloride reacts with Amommia & Ammonium chlende in the presence of activated charcoal to form tercamminecebaltiim chloride.

1. Chlsicle. In the above Eqn(reaction), rsquilpalmt of Cobalt (11) Ch Concle sooduses Equivalent of Herammine cobalt (m) chloride . The no. of miles of cobalt (11) chlesicle is Equal to na q miles of Hexammime cobalt (111) 1. for the theantallas yield calculation. . The no. q males of cobalt (1) chlsicle is = 7.5ml x 1.59 = 11.25 mmoles ". The theoritical yield of [co CNH3)6].el is = nog moes x m.wt-of (Co CNHz)s] che = 11.25 mmoles x 267.48 g mot = 3009.15 mg = 3.009159 Theoritical yield q [co (NH3)6]I3 = 3.009159. Given that actual yield g (Co (1H2)6] Uz = 7.82459. i percent yield= actual yield erex 100 thentical yield 2.82459 xlor 3. 009159 percent yield = 93.86%

palt Bl. Casoq. 5H20 + 4NH3 [CM (NH₃)]504 H2O + 4H₂O In Woods". Con Copper Sulphate pen ta hyecate reacts with Armonia to form te tra amine. copper (10) sulphate monohydlate. In the above realtim, mne Equivalent of Cu Soq5to produces (Equivalent of [cu CNH3)4]504: Hao. for calculating the theontical yield, one mole of Cu(SO4).5H2o - Imole of (au (N/H3)4] 504Hao . M. wt. q casoq:5420=249.689 /mol Miut.g [Cu (NH3 )47504' Hao = 245.74 g/mol 4. The mo. q moles of cusoq. Sh2o used = wt q Casoq. 54 mut. of Cu S04.5H₂O 3.50239 - 249.69 g molt = 0.014027 moles. q la so4.5H20 used. The thesitical yield of (cu (NH3)4]504 H2O a no-g meles x M.wt.of [Cu CNH3)47504. Hao = 0.01402 ** 245.74 gmort = 3.447 g. the critical yield of [cu (NH3)4] so4•H20 given actual yield = 299809

x100 - percent yield = - actual yield the sitical yield 29980 g 100 = 3.447 of casoq. Shao. percent yield = 86.97% Reasons for yield opel 100 Y:- . In pratically, the percent yield should be uss thom 100% But, However, the pergent Sields are greater tham 100% mostly due to the (1) measuled produced contains impulities which cause its mass greater thom the actual mals. (C) If the same side reactions takes place (ie) some times, the product can be wet con hygroscope deliquescent com probably effervescent. This will leads to the greater mats thom expected
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