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3. Coffee beans and tea leaves can be decaffeinated using ethyl acetate\\ ($CH_3CO_2C_2H_5$; density = 1.0492 g/mol). It is prepared by reacting\\ ethanol ($C_2H_5OH$; density = 0.7893 g/ml) with acetic acid ($CH_3CO_2H$;\\ density = 1.0492 g/mol); the other product is water.\\ a. If 10.00 ml of each ethanol and acetic acid are mixed, how many\\ grams of ethyl acetate can be expected to form? (6)\\ b. The reaction can be made faster if a small amount of sulfuric acid is\\ used as a catalyst, meaning it is not consumed in the reaction. The\\ reaction calls for 0.050 M sulfuric acid. If you only had a bottle of\\ 8.0 M sulfuric acid, describe in detail how would you prepare 75.0\\ ml of the required concentration of acid? (4)

          3. Coffee beans and tea leaves can be decaffeinated using ethyl acetate\\
($CH_3CO_2C_2H_5$; density = 1.0492 g/mol). It is prepared by reacting\\
ethanol ($C_2H_5OH$; density = 0.7893 g/ml) with acetic acid ($CH_3CO_2H$;\\
density = 1.0492 g/mol); the other product is water.\\
a. If 10.00 ml of each ethanol and acetic acid are mixed, how many\\
grams of ethyl acetate can be expected to form? (6)\\
b. The reaction can be made faster if a small amount of sulfuric acid is\\
used as a catalyst, meaning it is not consumed in the reaction. The\\
reaction calls for 0.050 M sulfuric acid. If you only had a bottle of\\
8.0 M sulfuric acid, describe in detail how would you prepare 75.0\\
ml of the required concentration of acid? (4)

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3. Coffee beans and tea leaves can be decaffeinated using ethyl acetate

(CH3CO2C2H5; density = 1.0492 g/mol). It is prepared by reacting

ethanol (C2H5OH; density = 0.7893 g/ml) with acetic acid (CH3CO2H;

density = 1.0492 g/mol); the other product is water.

a. If 10.00 ml of each ethanol and acetic acid are mixed, how many

grams of ethyl acetate can be expected to form? (6)

b. The reaction can be made faster if a small amount of sulfuric acid is

used as a catalyst, meaning it is not consumed in the reaction. The

reaction calls for 0.050 M sulfuric acid. If you only had a bottle of

8.0 M sulfuric acid, describe in detail how would you prepare 75.0

ml of the required concentration of acid? (4)

Added by Nathan F.

Chemistry: Structure and Properties

Nivaldo Tro 2nd Edition

Instant Answer

Solved by Expert Sri K

12/10/2022

Step 1

050 M sulfuric acid needed for the reaction. Given that the reaction calls for 0.050 M sulfuric acid and we need to prepare 75.0 ml of the required concentration, we can use the formula: M1V1 = M2V2 where M1 = initial molarity, V1 = initial volume, M2 = final Show more…

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Coffee beans and tea leaves can be decaffeinated using ethyl acetate ; density {(:=1.0492(g)/(m)ol)}. It is prepared by reacting ethanol ; density {(:=0.7893(g)/(m)l)} with acetic acid ; density =1.0492(g)/(m)ol; the other product is water. b. The reaction can be made faster if a small amount of sulfuric acid is used as a catalyst, meaning it is not consumed in the reaction. The reaction calls for 0.050M sulfuric acid. If you only had a bottle of 8.0 M sulfuric acid, describe in detail how would you prepare 75.0 ml of the required concentration of acid? (4) 3.Coffee beans and tea leaves can be decaffeinated using ethyl acetate (CH,CO,C,H; density = 1.0492 g/mol). It is prepared by reacting ethanol (C,H,OH; density = 0.7893 g/ml) with acetic acid (CH,CO,H; density =1.0492 g/mol); the other product is water. a. If 10.00 ml of each ethanol and acetic acid are mixed, how many grams of ethyl acetate can be expected to form? (6) b. The reaction can be made faster if a small amount of sulfuric acid is used as a catalyst,meaning it is not consumed in the reaction. The reaction calls for 0.050 M sulfuric acid. If you only had a bottle of 8.0 M sulfuric acid, describe in detail how would you prepare 75.0 ml of the required concentration of acid? (4)

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00:01 Dear friends, in the reaction between ascetic acidin 3 methyl 1butanol to form isopental acetate, the density is the volumes used and the molar masses of the deactansion products are given and you have to find out what is the mass of what is the volume of isopental acetate formed assuming the reaction goes to completion.

00:24 Now here first we will calculate the mass of acetic acidic acid and 3 methyl 1butanol used here.

00:33 So, mass of acetic acid, ascetic acid is mass equal to, here mass is not given, volume is given, so volume is 12 ml multiplied by density will give us mass, so mass we got 12 multiplied by 1 .05, divided by, if we divide by molar mass, we will get number of moles.

01:00 Here so number of moles will be here 12 5 zha so 0 .21 0 .21 mole of acetic acid similarly we'll find out moles of 3 methyl 1 butanol 3 methyl 1 butanol it is volume multiple moles equal to mass by molar mass mass is not given here also so it is 8.

01:33 Multiplied by density.

01:36 So we got mass here by multiplying volume with density.

01:40 Now divide with molar mass, 88 .1, we are getting the moles which is 0 .07 mole.

01:50 Now as for the equation here, 1 mole of acetic acid reacts with 1 mole of 3 methyl 1butanol to produce 1 mole of isopental acetate.

02:03 Now here, if we see here, acetic acid is 0 .21 mole whereas 3 methyl 1butanol is just 0 .07 mole.

02:13 Therefore, it is the limiting reactant.

02:16 3 methyl 1butanol is the limiting reactant.

02:23 Therefore, we will calculate the amount of isopental acetate formed based on this...

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