Sodium theoretical yields

It is evident from the equation that the sodium is used in excess. Actually 61 5 g. of n-propyl bromide is 0 5 gram mol this will react with 0 5 gram atom or 11-5 g. of sodium, so that 100 per cent, excess was actually employed. The theoretical yield of n-hexane will be 0-25 gram mol or 21-5 g., since 2 mols of n-propyl bromide give 1 mol of n-hexane. The actual yield was 18 g., hence the percentage yield is (18/21-5) X 100 = 84 per cent. 

The starting material for the above step may be prepared as follows 5 g (0.016 mol) of N -(p-methoxyphenyl)-p-chlorobenzhydrazide hydrochloride and 4.75 g (0.018 mol) of benzyl levulinoyloxyacetate were heated In 25 ml of glacial acetic acid for 3 hours at 80°C. The solvent was then evaporated off under vacuum. The residue was taken up in chloroform and the solution was washed neutral by shaking with sodium bicarbonate solution and thereafter with water. After drying the chloroform solution, this was subjected to chromatography on aluminium oxide, the eluate was concentrated by evaporation and the viscous oil remaining as residue was crystallized by adding ether. The compound melted at 94°-95 t. The yield was 4.1 g which corresponds to 50.7% of the theoretical yield. 

What is the theoretical yield of crystals which may be obtained by cooling a solution containing 1000 kg of sodium sulphate (molecular mass = 142 kg/kmol) in 5000 kg water to 283 K The solubility of sodium sulphate at 283 K is 9 kg anhydrous salt/100 kg water and the deposited crystals will consist of the deca-hydrate (molecular mass = 322 kg/kmol). It may be assumed that 2 per cent of the water will be lost by evaporation during cooling. 

Numerous methods for the synthesis of salicyl alcohol exist. These involve the reduction of salicylaldehyde or of salicylic acid and its derivatives. The alcohol can be prepared in almost theoretical yield by the reduction of salicylaldehyde with sodium amalgam, sodium borohydride, or lithium aluminum hydride by catalytic hydrogenation over platinum black or Raney nickel or by hydrogenation over platinum and ferrous chloride in alcohol. The electrolytic reduction of salicylaldehyde in sodium bicarbonate solution at a mercury cathode with carbon dioxide passed into the mixture also yields saligenin. It is formed by the electrolytic reduction at lead electrodes of salicylic acids in aqueous alcoholic solution or sodium salicylate in the presence of boric acid and sodium sulfate. Salicylamide in aqueous alcohol solution acidified with acetic acid is reduced to salicyl alcohol by sodium amalgam in 63% yield. Salicyl alcohol forms along with -hydroxybenzyl alcohol by the action of formaldehyde on phenol in the presence of sodium hydroxide or calcium oxide. High yields of salicyl alcohol from phenol and formaldehyde in the presence of a molar equivalent of ether additives have been reported (60). Phenyl metaborate prepared from phenol and boric acid yields salicyl alcohol after treatment with formaldehyde and hydrolysis (61). 

A. 81.37% is the percent yield. The question clearly notes that sodium hydroxide is the excess reagent. (Tip You always can ignore a reactant if the problem says it s in excess. That s like a big this-one-isn t-important sign in the problem.) So sulfuric acid is the limiting reagent and is the reagent you should use to calculate the theoretical yield ... 

The rapidity of the change of hypoiodites to iodates is illustrated by mixing 50 c.c. of a +A-soln. of iodine with 50 c.c. of a normal soln. of sodium hydroxide at 0°. The theoretical yield would be -AT-soln. of hypoiodite. In one minute the amount of hypoiodite in soln. is 5 per cent, less than the theoretical, and in two minutes 25 per cent. less. Less cone. soln. are a little more stable, for a A-soln. 

The maximum efficiency is clearly at the point Pi, for the potassium efficiency decreases in passing towards P3. A mixture containing 295 grass, of sodium nitrate, 24 5 potassium carbonate, and 50 grms. of water, after agitation, furnishes a soln. corresponding with the point Pi, and there separates 24 1 grms. of potassium nitrate contaminated with 4 5 per cent. of carbonate—the theoretical yield is 23 6 of potassium nitrate. 

Lead arsenate may also be obtained eleetrolytically by the anodic dissolution of lead in the presence of an arsenate. An almost theoretical yield has been obtained 4 by using a diaphragm cell with an anolyte containing 20 g. sodium arsenite and 70 g. sodium nitrate per litre and sufficient acetic acid for neutralisation, and a catholyte consisting of a 30 per cent, solution of sodium nitrate. The anode should be of lead and the cathode of iron, and the current density 55 to 6 amps, per sq. dm. 

The process of obtaining alkyl sulphonate in an autoclave with a mixer has been studied. Basic reagents were water solution of sodium hydrosulfite 36-38% and industrial olefin fractions at 240-320 °C. NaN03 and oxygen from air were used as initiators of the reaction of free radicals. System factors are x, reaction time, h x2 temperature of reaction, °C x3 mole ratio of sodium hydrosulfite and olefin x3 mole ratio of NaN03 and olefin x5 volume ratio of N-propanol and olefin. System response is a yield of alkyl sulfonate as a per cent of theoretical yield. FRFE design... 

The process of obtaining alkylsulfonate, used as raw material for synthetic products for washing, has been researched. Basic reagents are a water solution of sodium bisulfate and olefins. NaN03 and oxygen from air were used as initiators in reaction of free radicals. Based on previous studies, these factors were chosen xj-reaction time, h x2-reaction temperature, °C x3-mole ratio NaHS03 and olefins, [1] x4-mole ratio NaN03 and olefins, [1] and x5-volume ratio of N-propanol and olefins, [1], The yield of alkylsulfonate in per cent of theoretical yield has been determined as the... 

A slurry of sodium bicarbonate comprising 39.8 g sodium bicarbonate and 254 ml water was placed in an autoclave. 96.3 g hexadecyl bromide and 635 ml acetone were then added. The autoclave was sealed and while stirring (590 r.p.m.) it was heated to a temperature of 218°C over a period of 1 hour 15 min. The temperature was maintained at 218-220°C for an additional hour. At the end of the reaction the autoclave was cooled to about 50°C, that is, to a temperature at which the alcohol remains molten. The autoclave was then rinsed with acetone and 1 N hydrochloric acid add to neutralize the sodium bicarbonate. The reaction mixture was diluted with an equivalent volume of water and then extracted with n-pentane. (Other suitable water insoluble solvents such as benzene, carbon tetrachloride, chloroform, petroleum ether and the like can be used for extraction). The pentane extract was washed with water and then dried over magnesium sulfate. The dried solution was filtered and evaporated. The residue was melted and a vacuum applied to remove the last traces of pentane. On distillation a yield of 94.8% of the theoretical yield white crystals of hexadecanol was recovered M.P. 49°C, B.P. 344°C, nD79 = 1.4283. 

A solution of 34 cc (0.5 mol) of liquid phosgene in 150 cc of absolute ether is reacted while cooling with a mixture of sodium chloride and ice, first with 62 grams (0.5 mol) of 1-ethinyl cyclohexanol-1 and then with 64 cc (0.5 mol) of quinoline. The precipitated quinoline chlorohydrate is filtered off and the filtrate is reacted with ammonia in ether. In this manner 45 grams of the carbamic acid ester of 1-ethinyl cyclohexanol are obtained. Yield 53% of the theoretical yield. The ester boils at 108° to 110°C/3 mm and on recrystallization from cyclohexane, yields colorless needles melting at 94° to 96°C. 

A solution of 31.0 g (0.42 mol) of glycidol in 85 ml of anhydrous benzene and 46.0 g of sodium carbonate is heated to reflux. At this temperature and under a nitrogen atmosphere a solution of 70.0 g (0.27 mol) of crude 2,6,2, 6 -tetramethylbenzhydryl chloride in 120 ml of anhydrous benzene is added drop-wise. After completion of the addition, the mixture is kept refluxing for another 8 h. After filtration and removal of the solvent under reduced pressure, the yellow coloured residue is subjected to distillation under reduced pressure. There are obtained 64.0 g (80% of the theoretical yield) of a fraction boiling at 175-195°C/1 mm Hg, which slowly solidifies. This distillate, consisting of crude l,2-epoxy-3-[di-(2,6-xylyl)-methoxy]propane, is used for the next reaction step without purification. 

These three reactions can be used to calculate a theoretical yield of H2 processed per gram of sodium borohydride consumed. This calculation assumes that all necessary water generated in [A2] can be recycled for use in [Al]. Note that Reaction [A3] indicates that the process is a net producer of water, indicating borohydride is the only limiting factor. In a perfect recycle loop, no water would ever need to be added. 

The nitro compound layer is washed with warm water, then with a warm dilute solution of sodium carbonate said again with warm water. The washed nitro compound is dried by heating under reduced pressure. From 100 kg of p- chloronitrobenzene 128.5 kg of l-chloro-2,4-dinitrobenzene is obtained, which is 98.5% of the theoretical yields. 

The combination of the chemistry shown in Scheme 22,100 with the Sharpless kinetic resolution (SKR) of secondary allylic alcohols 46101 provides a method for the conversion of racemic allylic alcohols 46 into a single enantiomer with 100% theoretical yield.102 The reaction of sodium telluride with the mesylate 48 derived from 47 affords 46a. In this way, a single enantiomer of the allylic alcohol 46 is obtained in high yield (Scheme 23).102... 

Marie 2 was able to convert aconitic acid into tricarballylic acid at a mercury cathode surrounded with a solution of the acid half neutralized with sodium hydrate. Sixty per cent of the theoretical yield was obtained ... 

Often impure reactants are the cause of a percentage yield of less than 100%. Impurities cause the mass data to be incorrect. For example, suppose that you have 1.00 g of sodium chloride and you want to carry out a reaction with it. You think that the sodium chloride may have absorbed some water, so you do not know exactly how much pure sodium chloride you have. If you calculate a theoretical yield for your reaction based on... 

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