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Determining the Percent Yield of Calcium Carbonate

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Determining the Percent Yield of Calcium Carbonate Purpose: To compare the theoretical amount to the actual amount of calcium carbonate and calculate its percent yield in the reaction between solutions of sodium carbonate and calcium chloride. Materials: stirring rods fine filter paper 2 small beakers 0.8 - 1.2 g of sodium carbonate Erlenmeyer flask electronic balance graduated cylinder safety glasses 2 g of calcium chloride deionized water funnel retort stand ring clamp Safety Procedures: Sodium carbonate and calcium chloride are both harmful if swallowed or inhaled and may cause irritation to skin, eyes and respiratory tract. Safety goggles, and proper attire are to be worn at all times during this experiment. In particular, no loose clothing should be worn and long hair should be tied back. Procedure: 1. ...read more.


The filter paper was measured for mass. Results: Table 1: Masses of the Materials Material Mass (g) Filter Paper 1.77 Beaker 1 50.53 Beaker 1 with sodium carbonate 51.71 Beaker 2 101.46 Beaker 2 with calcium chloride 103.44 Filter paper with product 2.93 Table 2: Qualitative observations of the reagents before and after the reaction Course of time in the experiment Observations Before * sodium carbonate : white, powdery, odorless anhydrous salt that had a slower rate of dissolving in water and required extra water to completely dissolve; the solution was clear and transparent * calcium chloride: odouless white granules that dissolved easily in water; the solution was translucent During * as soon as the calcium chloride solution was poured into the sodium carbonate solution, a white, powdery precipitate was formed After * the residue in the filter paper was a white solid, powdery Calculations: Sample Calculation for determining ...read more.


Similarly, there are 1.11 � 10-2 moles of sodium carbonate in the experiment. NNa2CO3 < N CaCl2 Therefore, Na2CO3 is the limiting reactant. Number of moles of CaCO3 produced= Number of moles of Na2CO3 = 1.11 � 10-2 mol Mass of CaCO3= number of moles of CaCO3� molar mass of CaCO3 1 mol = 1.11 � 10-2 mol � 100.09 g 1 mol = 1.11 g Therefore, the theoretical mass of is CaCO3 1.11 g. Percent yield = experimental yield � 100% Theoretical yield = 1.16 � 0.02 g � 100% 1.11 g = 104.50 % Therefore, the percent yield is 104.50 %. Percent Error = | theoretical yield - experimental yield| � 100% theoretical yield = |1.11 g - 1.16 g| � 100% 1.11 g = 4.50% Therefore, the percent error in this experiment is 4.50%. ?? ?? ?? ?? - 1 - ...read more.

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