2. Abstract We weight the educts and products while we did the experiment, that we can see how the chemical substance comparability is balanced. 3. Introduction We mixed Na(2)CO(3) and HCl together. The reaction began and the product CO(2) came stunned as a muck up. The remnant was NaCl + H(2)O. 4. Procedure We install Na(2)CO(3) on the one side of the film over and the HCl(aq) on the other side. Then we took the funnel and stuck it in a safe and put both in the glass that no turgidity comes out. Then we took the piston chamber and filled it up with wet and put it upside down in a beaker unsounded of irrigate. There should be no turgidness in the cylinder! Now we shaked the glass with the two substances and the reaction began. There came bubbles (CO(2)) gas which went into the cylinder. Then we calculated the volume of gas in the cylinder. The irrigate was sided out after one day in the baker and we could also calculate how much water and NaCl in the residue was left. 5. Look on the paper behind. 6.         1) Na(2)CO(3) + 2HCl = 2CO(2) + NaCl + H(2)O 2) plentitude of Na(2)CO(3): 4.02g, 105.988g/ groyne         torpedo caboodle of NaCl: 58.443g/ breakwater         Mole Na(2)CO(3), 4.02g:105.988g/mole=0.0379mole         Mass of NaCl: 0.0379x58.443g/mole=2.215gx2= 4.43g                 3) Entire Mass: 5.05g                 NaCl: 86.65%                 H(2)O: 13.35%                 5.05gx0.8665= 4.375g                 4) difference= (4.43g-4.375g):(4.
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43g)x100                 difference= 1.24%                         5) Mol Na(2)CO(3) 4.02g:105.988g/mole=0.0379mole         22.4l/molex0.0379mole= 0.849liters of gas.         6) 680ml         7) difference=(849ml-680ml):849=0.199x100= 19.9% 7.         Conclusion I lettered how to balance a chemical equation in an experiment. I also proverb that the measured and calculated mass by gases very contrastive is, because we will lose all the cartridge holder a lot gas. If you want to get a climb essay, revision it on our website:
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