the concentration of sodium carbonate in an unknown sample by titration. The solution of hydrochloric acid was prepared and standardized using Na2CO3. Observations: Week 1: Standardizing hydrochloric acid using sodium carbonate with bromocresol green indicator Table 1: Titration of sodium carbonate using hydrochloric acid Trial | Mass of Na2CO3 (g) | Burette Reading (mL) | Final Volume of HCl (mL) | | | Initial | Final | | 1 | 0.2123 | 0.00 | 42.34 | 42.34 | 2 | 0.2195 | 0
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Analysis of Soda Ash and Volumetric Analysis of a Carbonate-Bicarbonate Mixture Buti‚ Mary Daphne A. Department of Chemical Engineering‚ Faculty of Engineering University of Santo Tomas España‚ Manila Abstract Sodium carbonate (Na2CO3)‚ commonly known by trade name soda ash‚ is a white‚ anhydrous‚ powdered or granular material that is an essential raw material used in the manufacturing of glass‚ detergents‚ chemicals‚ and other industrial products. In this experiment‚ the percent alkalinity
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In this practical the aim was to determine the concentration of Na2CO3 in a standard solution‚ by using Hydrochloric acid to neutralise it. This experiment involved using distilled water to dissolve the sodium carbonate until 250 ml of solution were made. I had to calculate‚ from the information provided‚ how much sodium carbonate to use in the making of the solution. Using the volume and concentration of the solution‚ I was able to calculate the moles of the solution (concentration x volume). Using
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between Na2CO3 and CaCL2. Sample Observations Na2CO3 solution COLORLESS CaCL2 solution COLORLESS Na2CO3 + few drops of CaCL2 Produced a cloudy‚ white mixture After Filtration Precipitate WHITE Filtrate COLORLESS C. FILTRATE TEST Table 4.5. The filtrate test. Condition Observations Filtrate + CaCL2 COLORLESS SOLUTION Filtrate + Na2CO3 FORMATION OF WHITE PRECIPITATE Table 4.6. Summary of the filtrate test. Group No. Color of filtrate Filtrate + CaCl2 Filtrate + Na2CO3 Limiting
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include determining the amount of Na2CO3 needed to do a full reaction. This was calculated through stoichiometry calculations: Molar mass was first calculated for CaCl2*2H2O Ca = 40.078g Cl2 = 35.453g*2 = 70.906g 2H2 = 1.00794g*4 = 4.03176g 2O = 15.9994g*2 = 31.9988g 40.078g + 70.906g + 4.03176g + 31.9988g = 147.01456g or 147.0 g CaCl2 1g CaCl2 * 2H2O x (1 mol CaCl2 *2H2O/147g CaCl2 *2H2O) = 0.0068 mol of CaCl2*2H2O Molar mass was then calculated for Na2CO3: Na2 = 22.9898g*2 = 45.9796g
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CaCO3 product on the labeled watch-glass. 17. Leave your labeled watch glass out for drying overnight (low temperature). 18. Record the mass of the dry filter paper + CaCO3. Record mass Questions and answers for Limiting Reactant 1. The mole of Na2CO3 used is n= m÷M = 3.01g÷105.99 =0.028mol 2. The moles of CaCO3 n= m÷M = 4g÷100.09g/mol =0.040 mol 3a: mass of CaCO3 produced is m=n×M = 0.028mol×100.09g/mol
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have to use stoichiometry. My calculation for the second reactant was: 1.0g CaCl2*2H2O(1mol CaCl2*2H2O/147g CaCl2*2H2O)(1mol Na2CO3/1mol CaCl2*2H2O)(106g Na2CO3/1mol Na2CO3) = 0.72g Na2CO3. The final‚ when its dried in the paper filtered weighed at 1.6 grams. Experiment and Observations: As I performed the lab‚ I had to retrieve all my equipment’s from my labpaq. As I started to setup‚ I had to weigh out 1.0 gram of CaCl2*2H2O on the digital scale. After it was weighed‚ it was added to the 100 mL
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|| || Data Tables: Step 3: Show the calculation of the needed amount of Na2CO3 Convert 1.0g of CaCl2-.2H2O to moles of CaCl2-.2H2O 1.0g x 1 mole CaCl2-.2H2O 147.0 g CaCl2-.2H2O = 0.00680 moles CaCl2-.2H2O The mole ratio is 1:1 Hence if we have 0.00680 moles of CaCl2-.2H2O we will as well need 0.00680 moles of Na-2CO3 Convert moles of Na-2CO3 to grams of Na2CO3 = 0.00680 moles Na-2CO3 x 105.99g Na-2CO3 1
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weigh 1g of CaCl2 +2 H2O‚ then pour it into the 100mL glass beaker with 25mL of distilled water and then stir it to make the solution calcium chloride. The next step was to calculate the amount of Na2CO3 I needed for it to completely react with the CaCl2 +2H2O using stoichiometry. 1.0g CaCl2+2H2O= 1mol CaCl2+2H2O =0.006802mol CaCl2+ 2H2O 147.0156g CaCl2+2H2O g of Na2CO3 = 105.98844g Na2CO3 0.006802 mol CaCl2+2H2O = 0.7209g Na2CO3 1 mol Na2CO3
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yield of the product from a precipitation reaction. Also‚ to learn concepts of solubility and the formation of a precipitate. Procedure: Weigh out your 1.0g of CaCl2-2H20 and put it into the 100mL beaker‚ Add your 25mL of distilled water and stir to form the calcium chloride solution. Next‚ use stoichiometry to determine how much Na2CO3 and put it into a small paper cup. Then add the 25mL of distilled water to make the sodium carbonate solution. Mix the two solutions in the beaker and a precipitate
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