2. Abstract We weight the educts and products while we did the experiment, that we can see how the chemical equation is balanced. 3. Introduction We mixed Na(2)CO(3) and HCl together. The reaction began and the product CO(2) came out as a gas. The residue was NaCl + H(2)O. 4. Procedure We put Na(2)CO(3) on the one side of the glass and the HCl(aq) on the other side. Then we took the funnel and stuck …showed first 75 words of 402 total…

…showed last 75 words of 402 total…gt;5) Mol Na(2)CO(3) 4.02g:105.988g/mole=0.0379mole <Tab/>22.4l/molex0.0379mole= 0.849liters of gas. <Tab/>6) 680ml <Tab/>7) difference=(849ml-680ml):849=0.199x100= 19.9% 7.<Tab/>Conclusion I learned how to balance a chemical equation in an experiment. I also saw that the measured and calculated mass by gases very different is, because we will lose all the time a lot gas.