Chemistry Investigation 14: ------------------------------------------------- To determine the enthalpy change of reaction for: ------------------------------------------------- Na2CO3(aq) + H2O(l) + CO2(g) → 2NaHCO3(aq) Given: S1— Anhydrous sodium carbonate (Na2CO3) S2— Anhydrous sodium hydrogen carbonate (NaHCO3) A1—Aqueous sulfuric acid (H2SO4)‚ 0.500mol dm-3 Apparatus | Uncertainty | Measuring cylinder | ± 0.5 ml | Electronic Balance | ± 0.001 g | Data logger | ±0.2 ℃ |
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February 27‚ 2013 Abstract The analyte used is the soda ash which is titrated with an HCl titrant‚ standardized by Na2CO3. The indicators used are phenolphthalein for basicity and methyl orange for acidity. The two volumes of the titrant are then used to calculate percent composition of soda ash analyte. At the end of the experiment‚ the calculated average percent by mass of Na2CO3 is 4.92% and the average percent by mass of NaHCO3 is 5.07%. Introduction Soda ash is the common name for sodium carbonate
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Treatment (3) Environment Impact Research After hours of vigorous examination‚ we have come to the conclusion that calcium chloride (CaCl2) would be the best deicer to use on a naval base over sodium chloride (NaCl) and ethylene glycol (C2H6O2). In order to melt ice‚ deicers lower the freezing point so that the ice melts at lower temperatures. CaCl2 has a freezing point depression of 5°C‚ whereas NaCl has only a 3°C depression‚ and ethylene glycol‚ 1.85°C. The more the freezing point can
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mole NaOH) / (1 mole CaCl2) ]*[ .005 mol CaCl2 ] = .010 moles NaOH needed to completely react CaCl2‚ only .005 moles available‚ therefore NaOH limits the reaction Theoretical Yield of Ca(OH)2: [ .005 moles NaOH ] * [ (1 mole Ca(OH)2 ) / (2 moles NaOH) ] = .0025 moles Ca(OH)2 formed [ .0025 moles Ca(OH)2 ] * [ (74.0926 g Ca(OH)2 ) / ( 1 mole Ca(OH)2 ) ] = .1852 grams Ca(OH)2 Percent Yield [ .25 g / .1852 g ] * 100 % = 134.989 % Test #2 Volume CaCl2: 5.0 mL = .005 L
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GROUP B MODEL ANSWER Howard et al.‚ Temperature-induced structural changes in CaCl2‚ CaBr2 and CrCl2: A synchrotron X-ray powder diffraction Phys. Rev. B 72‚ 214114 (2005). DO NOT PLAGIARISE THIS MODEL ANSWER Paragraph 1 The polymorphs involved are the halides in tetragonal rutile (TiO2) structure and the halides in so-called calcium chloride structure. As the name implies‚ the unit cell of halides in tetragonal rutile structure is tetragonal‚ with the lattice parameters a=b≠c and α=β=γ=90o
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using Method of Double Indicators Student Handout Purposes To determine the composition of the following mixture by double indicator method: 1. NaOH(aq) and Na2CO3(aq) 2. NaHCO3(aq) and Na2CO3(aq) Introduction Consider a mixture of NaOH(aq) and Na2CO3(aq). Reaction between HCl(aq) and Na2CO3(aq) takes place in two stages: HCl(aq) + Na2CO3(aq) ⎯→ NaHCO3(aq) + H2O(l) …………………. (1) HCl(aq) + NaHCO3(aq) ⎯→ NaCl(aq) + CO2(g) + H2O(l) …………. (2) While that between HCl(aq) and NaOH(aq) completes in only
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Heating 2NaHCO3→Na2CO3+CO2+H2O a) Experimental results i) Trial 1 Mass of NaHCO3 | 0.3g | Temperature of hot plate | 400 ˚C | Time interval | 10 min. | Mass of mini beaker and product (Na2CO3) | 31.3g | Mass of mini beaker | 31.21g | Final mass of Na2CO3 | 0.09g | ii) Trial 2 Mass of NaHCO3 | 0.3g | Temperature of hot plate | 400 ˚C | Time interval | 10 min. | Mass of mini beaker and product (Na2CO3) | 31.52g | Mass of mini beaker | 31.31g | Final mass of Na2CO3 | 0.21g |
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2(0.1002±0.0002)(0.999)(0.04050±0.00005)(105.99)=0.0467±0.0012 AverageMHCl= 0.0424 ±0.0114+ 0.0420 ±0.0113+ (0.0467 ±0.0112)3=0.0437 ±0.0009 %Na2CO3=(Ave. MHCl)(Vph)(MW Na2CO3)g samplex100% %Na2CO3(1)=(0.0437)(0.004 ±0.00005)(105.99)0.0998 ± 0.0002x100%= 18.56 %± 0.0127% %Na2CO3(2)=(0.0437)(0.0041 ±0.00005)(105.99)0.0998 ± 0.0002x100%= 19.06 %± 0.0122% %Na2CO3(3)=(0.0437)(0.003 ±0.00005)(105.99)0.1002 ± 0.0002x100%= 13.87 %± 0.0167% %NaHCO3(3)=0.0437[0.00980±0.0005-(0.0030±0.00005)](105.99)0.1002±0
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Pallagani Investigating Stoichiometry with sodium salts of carbonic acid 4/2/14 A summary of the concepts: The purpose of this lab is to better understand “stoichiometry”. We will be reacting sodium carbonate (NaHCO)3 and sodium carbonate (Na2CO3) with hydrochloric acid to produce sodium chloride‚ water‚ and carbon dioxide. The balanced chemical reaction looks like this: NaHCO3 + HCl = NaCl + H2O + CO2 CAUTION: Be especially careful when handling the 6M HCl (aq)‚ as it can cause chemical
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HCl with Sodium Carbonate: Na2CO3 + 2HCl 2NaCl + H2O + CO2 (2) Concentration of Sodium Carbonate (Na2CO3) Molar mass of Na2CO3 : 22.99*2+12.01+16*3=105.99 Mass of Na2CO3 : 1.3 g Number of moles : n=mM =1.3 105.99=0.012265308 mol Volume Na2CO3 solution : 250 mL : 0.250 L Concentration: Concentration of Sodium Carbonate is 0.049 mol L-1 c=nV = 0.012265308 0.250=0.049061232 mol L-1 (3)Concentration of HCl Volume of Na2CO3 solution : 20 mL : 0
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