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    yield. Make sure to show all stoichiometric calculations and all data in your lab report. Na2CO3(aq) + CaCl2. 2H2O(aq) → CaCO3(s) + 2NaCl(aq) + 2H2O Data‚ observations‚

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    Almansoori Lab 17

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    Table 3: Comparison of theoretical and actual yields for CaCO3 Trial # Limiting Reagent Theoretical Yield of CaCO3 Actual Yield of CaCO3 % Yield Trial 1                   Trial 2                         Trial 3                         2. Calculate the theoretical yield of CaCO3 that could be produced by each trial and then fill in Table 3.       3. Find the percent yield each trial obtained for the CaCO3.       Pre-Lab Questions 1. A limiting reagent is one that

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    Experiment 9 okiemute

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    Name: Exp 9: Stoichiometry of a Precipitation Reaction Data Tables: Step 3: Show the calculation of the needed amount of Na2CO3 CaCl2.H2O(aq)= m/M =1/147 =0.0068 mol CaCO3(s)=0.0068*1/1 =0.0068 mol CaCO3 (s)= CaCO3 mol *CaCO3 g =0.0068 mol*100.01 g =0.68 g Convert moles of Na-2CO3 to grams of Na2CO3 = 0.00680 moles Na-2CO3 x 105.99g Na-2CO3 1 mole Na-2CO3 = 0.72g 0.72g of Na-2CO3 to fully react with 1g of CaCl2-.2H2O Step 4: Mass of weighing dish

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    they need to neutralise the soil for they’re farming. Limestone is commonly used for neutralising soil; it is a sedimentary rock that is commonly found in quarries. It is a compound consisting of calcium‚ carbon and oxygen called calcium carbonate (CaCO3). As it is a solid it would be very hard to neutralise soil with it so it has to be broken down and made into a liquid form called calcium hydroxide (CaOH2) made up of calcium‚ carbon‚ oxygen and hydrogen. This calcium hydroxide can then be sprayed

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    the calculation of the theoretical yield of calcium carbonate. The mole ration between CaCl2-.2H2O and CaCO3 is 1:1 that means that if we have 0.00680 moles of CaCl2-.2H2O we will get 0.00680 moles CaCO3 Convert the moles of CaCO3 to grams of CaCO3 = 0.00680 moles CaCO3 x 100 g CaCO3 1 mole CaCO3 = 0.68g CaCO3 Show the calculation of the percent yield. = Actual yield/Theoretical yield x 100 = 0.5/0.68 x

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    Chemistry notes

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    complex-ions: Cu(H2O)42+ (aq) + 4Cl־(aq) CuCl42־(aq) + 4H2O (l) Decomposition of dinitrogen tetroxide: N2O4 (g) 2NO2 (g) Decomposition of calcium carbonate: CaCO3 (s) CaO (s) + CO2 (g) Non-Arrhenius acid/base reaction: Gaseous hydrogen chloride and ammonia react: HCl (g) + NH3 (g) NH4Cl (s) IONISATION of strong and weak acids: Hydrochloric: HCl (g) + H2O (l) H3O+ (aq) + Cl־ (aq) Nitric: HNO3 (l) + H2O (l) H3O+ (aq) + NO3- Sulfuric: H2SO4 (l) + 2H2O (l) 2H3O+ (aq) + SO42־ Ethanoic: CH3COOH

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    CaCl2·2H2O and record in the table. (See lab‚ page 79 for example) Prepare a table. 3. Calculate the number of grams of Na2CO3 to weigh out from the number of moles in step 2. Prepare a table. 4. Calculate the number of grams of CaCO3 that are expected to be produced. This is your theoretical yield. Have Factor (molecular mass) Factor (mole ratio)

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    acid-base neutralization titrations were conducted for experimental analysis. The first set of titrations was to standardize a solution manufactured in the lab. An approximate solution of Na2EDTA of 0.004 M was titrated against a known solution of 1.000 g CaCO3/L to deter mine to exact molarity of the Na2EDTA. Ca2+ + Na2EDTA → CaEDTA + 2Na+ The second set of titrations was to use the now standardized Na2EDTA solution to determine an unknown water sample’s hardness. The unknown water sample is # 89. Water’s

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    Soap Experiment

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    Abstract An experiment is conducted to prepare soap and thus‚ to compare the properties of the prepared soap and synthetic detergents which are precipitation‚ emulsification and cleaning abilities. It can be concluded that soap has the properties if emulsifying oil whereas detergent has not. The abilities of forming precipitates can be seen clearly in soap solution whereas detergent forms no precipitates at all. The experiment is completed and successfully conducted. Introduction

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    A Paper

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    metal carbonate‚ M2CO3. There are 3 main reactions in this lab: 1. Equation 1: M­2CO3 (s) → 2M+ (aq) + CO32- (aq) 2. Equation 2: Ca2+ (aq) + CO32-(aq) → CaCO3 (s) 3. Equation 3: CaCl2 (aq) + M2CO3→ CaCO3 (s) + 2MCl (aq) The precipitated calcium carbonate is then filtered‚ dried‚ and weighed. The moles of calcium carbonate‚ CaCO3‚ are equal to the moles of Group 1 metal carbonate‚ M2CO3‚ added to the original solution. Dividing the mass of the unknown carbonate by the moles of calcium carbonate

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