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Dry the combined dichloromethane extracts with anhydrous sodium sulfate. Then decant this dried solution into small, tared beaker equipped with boiling chip. Rinse ...

Question

Dry the combined dichloromethane extracts with anhydrous sodium sulfate. Then decant this dried solution into small, tared beaker equipped with boiling chip. Rinse the sodium sulfate once with dditional dichloromethane ad add this rinse t0 the tared beaker. Remove the dichloromethane by evaporation in the sand bath to obtain Product A_ This product is sufficiently pure as isolated to allow for experimental yield calculation and characterizationTransfer the aqucous phase from the separatory funne

Dry the combined dichloromethane extracts with anhydrous sodium sulfate. Then decant this dried solution into small, tared beaker equipped with boiling chip. Rinse the sodium sulfate once with dditional dichloromethane ad add this rinse t0 the tared beaker. Remove the dichloromethane by evaporation in the sand bath to obtain Product A_ This product is sufficiently pure as isolated to allow for experimental yield calculation and characterization Transfer the aqucous phase from the separatory funnel to a 150-mL beaker and heat it in the sand bath for several minutes tO remove any traces of dichloromethane. Tnen cool the aqueous phase in @n ice- water bath and acidify it With 3 M hydrochloric acid. Swir] the mixmure during acidification_ Collect the precipitated product (which will be called Product B by vacuum filtration. Then wash the crystals with several 3-mL portions of cold, deionized water, followed by 2 ml of cold ethanol. Air-dry Product B until subsequent laboratory period. Then recrystallize this Product _ from small amount of 959 ethanol, and collect the crystalline product by vacuum filtration with Hirsch funrel. Rinse the crystals with 2 mLof cold 9590 ethanol and allow them to air dry. Once recrystallized, Product B sufficiently PuTe to allow for experimental yield calculation and characterization: Experimental Yield and Characterization Weigh products A and B_ and calculate experimental yields _ Tnen determine the melting ~poini range for each product and compare the observed values with the literature values for benzamide and benzoic acid Perform mixed melting ~point = determinations of cach product with authentic benzamide and benzoic acid t0 idenlify Product and Product B Finally. submit the two products in properly labeled product bags with the report



Answers

In the final stage of the manufacturing process for a solid organic product, the product is cleaned with liquid toluene and then dried in a process whose flowchart is shown on the next page.The wet product enters the dryer at a rate of $300 \mathrm{Ib}_{\mathrm{m}} / \mathrm{h}$ containing $0.200 \mathrm{lb}_{\mathrm{m}}$ toluene $/ \mathrm{lb}_{\mathrm{m}}$ dry solids. A stream of nitrogen at $200^{\circ} \mathrm{F}, 1.2 \mathrm{atm},$ and containing a small amount of toluene vapor also enters the dryer. (A higher temperature would cause the product to soften and degrade.) Heat is transferred in the dryer from the gas to the wet solids, causing most of the toluene to evaporate. The final product contains $0.020 \mathrm{lb}_{\mathrm{m}}$ toluene $/ \mathrm{lb}_{\mathrm{m}}$ dry solids. Gas leaves the dryer at $150^{\circ} \mathrm{F}$ and 1.2 atm with a relative saturation of $70 \%$ and passes through a water-cooled condenser. Gas and liquid streams leave the condenser in equilibrium at $90^{\circ} \mathrm{F}$ and 1 atm. The gas is reheated to $200^{\circ} \mathrm{F}$ and reenters the dryer.(a) Briefly explain this process in your own words. In your explanation, include the purposes of the condenser and the nitrogen reheater and a likely reason that nitrogen rather than air is used as the recirculating gas. What do you suppose happens to the liquid toluene leaving the condenser?(b) Calculate the compositions (component mole fractions) of the gas streams entering and leaving the dryer, the circulation rate of dry nitrogen ( $\left(\mathrm{b}_{\mathrm{m}} / \mathrm{h}\right),$ and the volumetric flow rate of gas entering the dryer (ft $^{3} / \mathrm{h}$ ). (c) Explain why the actual process has a small make-up nitrogen stream blended with the feed to the blower.

So a solution of magnesium sulfate helped A hydrate on insoluble solids is heated to resolve the MBS so far in water and then filtered to remove are solid, followed by a cooler that causes re crystallization in a second field to step to remove the crystals. I'm making recycle felt trade from this back into the system. So in the first part of the podcast, we will be discussing that the heating of the solution in the tank dissolves all of the M G s 04 on top seven inch, 20 on. Then filtering removes the insoluble minerals in the saturated solution cooling by we crystallizes MGs so far which could be recovered by filtering again. So moving on to the second part here. So when is calculate the mass of filter cake out followed. I'm dark cake. 15 I'll be, um, cake divided by one. Uh, l b m asked. Multiplied by excess. Well, stop. I am talk or so. What we get is value off 6300 LV. Put our. So next we can write the mass violence equation around the entire system for the M. G s 047 h 20 following this. Once we have, we have a mass fraction of MGs 047 h 20 in or that it's not 200.9. So the ability of saturated magnesium 110 F is not going 32 and a 50 F. It's not going to three on the mass of the crystal for £5 saturated solution at 50 F is £100 so he substitute these values in on. Then, following this, we got em dot c that is 55 194 l b and our so following this because it'll number amount of the cake from the second field so we can calculate the production rate of the crystal product by cutting for the £5 of solutions £100 for solid. So we have dark crystal people toe see, which we just calculated by by 1.5 I'll be cake over. I'll be, um, crystal. We could value off 52,000 565 l b per hour. So the flurry of the water we confined by doing a total balance around the system, so m dot w secret to 4 1009. 4 l b a hour. So, looking at the final part now we can calculate the ratio of the recycled master water max, which is off uh, W people to m dot off. Um, Doc W. That's equal to 95.795 He'll be Emperor Tried that by 1494 I'll be I'm proud. About 64 pounds cycle, pa pound of water.

So I am. Exchange chromatography is executed by making the reversible exchange of irons between irons present and solution on iron exchange resin. So we tend to use this to separate similar irons. And so in the first part where it's calculating the total amount of solution fed into the column 36 7.2 liters. So in part B, the expression of a volumetric flow rate for solution a be a T think was not point for liters per minute. The VA 33.6 minutes multiplied by t was able to not point not 11 90 so we can apply the mass balance in solution A M put a is equal to accumulation at a and we are able thio. The salmon that has Time T varies from not to 33.6 minutes and V A varies from zero to V a bracket. L The volume of Solution A becomes 6.72 liters on. We can calculate the amount of sodium chloride there's an A. C. L is equal to 392.72 grams and a c l. And so we can give an expression for environment for a lower rate of solution be that is tu minus no point no. 11 19 that is equal to be be dark t. So the total volume of solution a 60.48 liters total volume solution by solution be not at all number of moles of tres fad. Three points. There are two full mall on some possible potentials for error. So pumped be might have failed, causing a flow of one mole of sodium chloride to the column during the illusion period. Programs method chosen to execute in the chromatography could be incorrect. Perhaps the column was packed with the wrong reason, maybe arising. That has reduced affinity compared to the correct resin.


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