94 5.Limiting Reactant Synthesis of Nicke0) Complex Stoichiometric ratio: NiCl,

Question

94 5.Limiting Reactant Synthesis of Nicke0) Complex Stoichiometric ratio: NiCl, 6H,O t Ni(H,NCH,CH,NH),Cl, Stoichiometric ratio of H,NCH,CH,NH, to Ni(H,NCH,CH,NH),CI, Moles of NiCH,NCH,CH,NH),CI, if NiCI, 6H,O 0. 011074 No is the limiting reactant is the limiting reactant Identity of the limiting reagent Mass of NiCl, 6H O Moles of Ni(H,NCH,CH,NH),CI, if H,NCH,CH,NH, .0250 mo consumed during the reaction Mass of H,NCH,CH,NH 10 consumed during the reaction Calculated grams of excess reactant Molar mass of Ni(H,NCH,CH,NH)C, Theoretical yield (in mass) of Ni(H,NCH,CH,NH),CL, 12 produced Theoretical yield (in mass) of H,O produced Total mass of all reactants Total mass of all products

Explanation / Answer

The reaction for the experiment is:

NiCl2 + H2NCH2CH2NH2 ------> Ni(H2NCH2CH2NH2)Cl2

Moles of Ni(H2NCH2CH2NH2)Cl2 formed is more when H2NCH2CH2NH2 is limiting reagent.

Thus, NiCl2.6H2O is limiting reagent.

Mass of NiCl2.6H2O consumed during the reaction:

Moles of Ni(H2NCH2CH2NH2)Cl2 formed when NiCl2.6H2O is limiting reagent = 0.011074 mol

molar mass of NiCl2.6H2O = 237.7 g/mol

mass of NiCl2.6H2O consumed = moles*molar mass = 0.011074*237.7 = 2.632 g

molar mass of H2NCH2CH2NH2 = 60.2 g/mol

Moles of H2NCH2CH2NH2 consumed = moles of NiCl2.6H2O consumed = 3(0.011074) mol

Mass of H2NCH2CH2NH2 consumed = 3(0.011074)*60.2 = 2.000 g

Moles of excess reagent actually present = 3(0.0250) = 0.0750 mol

moles of excess reagent left after reaction = 0.0750 - 3(0.011074) = 0.0418 mol

mass of excess reagent = 0.0418*60.2 = 2.516 g

molar mass of Ni(H2NCH2CH2NH2)Cl2 = 189.7 g/mol

Theoretical yield of Ni(H2NCH2CH2NH2)Cl2 produced = 0.011074*189.7 = 2.101 g

Theoretical yield of water produced = 6*0.011074*18 = 1.196 g

total mass of all reactants = 2.632+2.000 = 4.632 g

total mass of all products = 2.101+1.196 = 3.297 g

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