For the table in #7, how do you calculate the moles of the 3 substances and also

Question

For the table in #7, how do you calculate the moles of the 3 substances and also what is the amount of 2-chloro-2-methylbutane? I finished the rest, I only need those 4 boxes.

in water. Excess acid acid and an crude product by wash washing the product insoluble in with by wilare removed by The the sodium? sueateeous sodium chloride and dry Trace amounts of te Trace base, sodum bicartorate purified Aranted using distillation to remove any other wuct 12-Chlor-2-methylbutane) and starting mater/ . nnydrous sodium surate. The crude product wil be be analyzed by IR spectroscopy ane) will be Prolab road the following 5. Read the entire experiment, review sing eparatory sections from your techniques sector ques section: separatory funnel and infrared review single distillation and 6. Complete title, date of experime Complete tion tableiseriment,purpose, procedure and data tables. A elo titaution table is s required 7. Complete the shaded areas of areas of the reagent and product tables, but make your own table. Shature prett-2-butanol drocheneAod 2-methyl-2-butanol 2-Chloro-2-methylbutan OH H-Cl Cl M" Molar ty MaBo Moles 12M Density 10.0 mL 25.0 mL Molar Mass ** Theontically 8. Based on the above information, what is the limiting reagent of the experiment? above i 9, What is the WHIMS symbol associated with concentrated HCl? eagent

Explanation / Answer

Answer:

The number of moles of substance is calculated by formula

Moles = (Weight of substance / Molecular weight of subtance) * Purity

e.g. consider 4gm of 2-methyl 2- butanal is taken for reaction having purity 99.2% (purity and molecular weight mention on substance bottle) then moles becomes

moles= ( 4 gm/ 88.15) *0.992

Moles = 0.045

similar way you can calculate moles of 3 subtance.

theoretical amount of 2- chloro-2-methyl butane is calculated from reaction

2-methyl-2-butanol + HCl ------------------> 2- chloro-2-methyl butane

MW-88.15 MW-106.59

88.15 gm of reactant forms 106.59gm of product

then Xgm of reactant forms how many gms?

(e.g. if we take 4 gm of reactant then amount of 2- chloro-2-methyl butane is

88.15 gm -----------. 106.59 gm

4 gm ------------?

= (4X106.59)/88.15

=4.83 gm )

I hope you will understand properly. Similarly you can calculate for your questions.

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