Question

H.Y. & S. D. Precipitating Calcium Carbonate CHMI21 General Chemistry Lab 1 • Treatment of Data: 14) mole amount of Na CO, 15) mole amount of CaCl 16) limiting reagent .00832 00976 NonCO₃ .00735 .832 .au 112.98 mole amount of CaCO, 17) 18) 19) 20) Theoretical Yield (g) of Cacos Actual Yield (g) of Caco, Percent Yield (%) Calculations (Show your work here): 14) the number of moles of Na:CO; used mol F 1704M 80.000L = 0.00832 15) the number of moles of CaCl, used mol = 1.22M X 0.008L = 0.00976 16) Identify the limiting reagent in this experiment based on the calculations in 15) and 16) Naz LOs because it is the smaller of the two 17) Calculate the theoretical yield of CaCO3 in moles 20) Calculate the percent yield

I just need help understanding the steps to get the theoretical yeild and percent yield of CaCO3. using the given information.

Answer 1

walanced equation: Cacta + Na, cog lan) (aq) Identification of Cacozt & pace 15 49) limiting Reagent Holarity weigar y looo rout volute (My rolarity & tiwtX Volume weight = logo weight of Taug = 1.29 M X 110.98 x 8M41 IMG 10oo in = 1.083 9h8 roles of calles, weightline) MocIt - 1093 110.98 = 0.00976 Moles lob 9.76 7. roles Holes weight of Nagcozz 1.04HX 8tiX 106 Tooo yns = 0,881 9M6 roles of Nag coz = weight = 0.821 gros Citrole = looo Potos) 11.832 holes (ob 8.39 hitoves anong the two Reactants, Nas Coz has 1888 Numbered mes to say the Rosen H.94 lot - OCR

calculation of theoretical yield (loot yield): 106 gMg of Nalog produced too grig of Calog 0.881 grs of NgCoz produces ? gtig of caces = 0.881 X 100 grs of Caloz Theoretical yield 2 10.831 gry of Calog of of yield calculation; % of yield Experiental weight. - X 106 theoretical weight 3 0.94 X100 = 113.117 J. of yilld = 113 11 Note: always w generally for any reaction the it. of yield 100% con 100% below. But here we got at indicated It is Impore cacozo