How to balance a chemical equation, shown in an experiment, by Mathias Zollbrecht

2. Abstract

We weight the educts and products while we did the experiment, that we can see how the chemical comparability 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 muck up. The remnant was NaCl + H(2)O.

4. Procedure

We put 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 rubber and put both in the glass that no gas comes out. Then we took the piston chamber and filled it up with wet and put it upside down in a beaker sound of irrigate. There should be no gas 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 water 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/mole

        Molar caboodle of NaCl: 58.443g/mole

        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.



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 saw that the measured and calculated mass by gases very different is, because we will lose all the cartridge holder a lot gas.

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