Stoichiometry Lab Report

Satisfactory Essays
Hugh Kim

Lab Report: Stoichiometry Lab

1. Prelab

Part1.

1) Create no waste = The principle that encourages chemists to not create waste at the first place rather than cleaning it up afterwards effectively shifts the chemistry more environmentally conscious, as creating no waste would make the experiment efficient; the reactants will be reduced to only the essential ones and the product will be maximized, a change that would make the experiment economic. Also, if chemists aim to engender no waste, this principle would solve the problem of disposal; as waste is often disposed by throwing it away in the nature, less waste possible will inflict less harm to the environment. Also, sometimes, the left over from the experiment
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Peer review is a process of rendering other people with the knowledge of the subject to check the process and the result that the experimenter has yielded. Thus, receiving peer reviews before publication is significant, as peers with different perspectives can point out and inform the experiments about the errors, any possible source of obfuscation, misleading parts and parts that the experimenter might have not considered. It also generally increases the quality of the report after the discussion with peers. Thus, chemists whose process could be propagated and could be used by others, wrong instructions could lead people to cause mistakes, and dangerous accidents. Thus, to reduce those risks, peer review could offer great benefits of making the experiments more secure. In essence, the results without any peer reviews has a higher chance of containing errors, a phenomenon that should be prevented before reporting the results of chemical …show more content…
In step1, one set up x as a mass of NaHCO3 in the original mixture. Thus, equation that shows that the total mass of NaHCO3 and Na2CO3 is 2.00g. In the second step, by using gravimetric analysis, one converts x into the mass of Na2CO3 that has been produced from heated NaHCO3, and states that the total mass of produced Na2CO3 and the Na2CO3 in the original mass is 1.605g, the result that the experimenter gained from the experiment. In the step4, one figures out the x, the mass of NaHCO3 is 1.070g. In step5, the percentage of NaHCO3 is being calculated by using the number that one gained from step4.

2NaHCO3(s)  Na2CO3(s) + CO2(g) + H2O(g)

1) 2.00g = xg NaHCO3 + (2.00g –xg) Na2CO3
2) According to the equation above, xg NaHCO3 X (1mol NaHCO3/84.01g NaHCO3) X (1mol Na2CO3/2mol NaHCO3) X (105.99g Na2CO3/1mol Na2CO3) + (2.00g – xg NaHCO3) = 1.605g –(a)
3) 0.6308x + 2.00g – x = 1.605g
4) x = 1.070g NaHCO3  --(b)
5) %NaHCO3 = (1.070g/2.00g) X 100% = 53.50%

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