a. Determine which of the reactants was thelimiting reactant, which was in exces

Question

a. Determine which of the reactants was thelimiting reactant, which was in excess and calculate the%

yield.

b. Howmany grams of each reactant will be left over at the end of thereaction?

Iwould appreciate any help I can get, I am stumped!

In a sample experiment, 1.0 g of CaCl2•H2O was reacted with 1.0 g of K2C2O4 • H2O. Afterdrying, 0.60 g of CaC2O4 •H2O were isolated.

a. Determine which of the reactants was thelimiting reactant, which was in excess and calculate the%

yield.

b. Howmany grams of each reactant will be left over at the end of thereaction?

Explanation / Answer

   CaCl2.H2O +K2C2O4.H2O -->CaC2O4.H2O + 2KCl +H2O    No. of moles of CaCl2.H2O = mass/molarmass = 1/129 moles = 0.00775 moles    No. of moles ofK2C2O4.H2O = 1/184moles = 0.00543 moles       Clearly, the limiting reagent isK2C2O4.H2O andCaCl2.H2O is in excess    No. of moles of product formed = 0.00543 moles       Mass of product formed = 0.793 gms       Actual yield = 0.6 gms       % yield = 0.6/0.793 x 100% = 75.66%    b) No. of moles of product formed = 0.6/146 = 0.0041moles    Mass of CaCl2.H2O remaining =(0.0078-0.0041) x 129 moles = 0.476 gms    Mass of K2C2O4.H2Oremaining = (0.0054-0.0041) x 184 moles = 0.239 gms

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