Percentage Yield Experiment
Where n is the number of moles, m is the given mass of a substance in grams, and M is the molar mass of …show more content…
When both solutions are mixed a white precipitate (Calcium Carbonate) is formed alongside Sodium Chloride. A clear colourless liquid is observed as the white precipitate is filtered through the filter paper. …show more content…
Amount of CaCl2 = m/M = (0.5g )/( 110.98 g/mol ) = 0.004505316 mols of CaCl2.
Amount of CaCO3 = 0.004505316 x 1/1 = 0.004505316 mols of CaCO3.
Molar mass of Na2CO3 = 2(22.99 g/mol) + 12.01 g/mol + 3(16.00 g/mol) = 105.99 g/mol.
Amount of Na2CO3 = m/M = (0.7g )/( 105.99 g/mol ) = 0.006604396 mols of Na2CO3.
Amount of CaCO3 = 0.006604396 x 1/1 = 0.006604396 mols of CaCO3.
Thus the Calcium Chloride is the limiting reagent as it produces less Calcium Carbonate.
Molar Mass of CaCO3 = 40.08 + 12.01 + 3(16.00) = 100.09 g/mol of CaCO3.
Theoretical yield of CaCO3 = M x n = 100.09 g/mol x 0.004505316 mols = 0.450937078g of CaCO3.
Actual yield of CaCO3 = Mass of product and filter paper - Mass of dry filter paper = 1.02g – 0.59g = 0.43g of CaCO3.
Percentage Yield = (0.43g )/( 0.450937078g ) x 100% = 95.35698459 % 95%.
Amount of used Na2CO3 = amount of limiting reagent x mole ratio = 0.004505316 x 1/1 = 0.004505316 mols of Na2CO3.
Amount of excess Na2CO3 = 0.006604396 - 0.004505316 = 0.00209908 mols of