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If 2.03 g of salicylic acid is reacted with 61 mL methanol, what is the maximum

ID: 879112 • Letter: I

Question

If 2.03 g of salicylic acid is reacted with 61 mL methanol, what is the maximum mass of product (oil of wintergreen) that could be formed? The density of methanol is 0.791 g/mL. Give your answer in grams with 3 significant figures.

If 2.03 g of salicylic acid is reacted with 61 mL methanol, what is the maximum mass of product (oil of wintergreen) that could be formed? The density of methanol is 0.791 g/mL. Give your answer in grams with 3 significant figures.

Question1 Match the following acids and alcohols to their corresponding esters Question HCOOH+ CH3(CH2)2CH20H> B. CH3COOCH2CH2CH3 CH3COOH + C6H5OH E. CH3CH2CH2COOH + CH30H-> c. CH3CH2CH2CooCH3 + H20 CH3CH2COOH + C6H50H A. CH3CH2Co0C6H5+ H20 CHacooH+CHaICH2 20HCHBCH2CH2COocH3+ H20 Selected Match Question 2 Name the esters formed in Question 1. (Dont worry about the order. Ignore Water) [Blank 1] [Blank2] [Blank3] [Blank4][Blank5] Specified Answer for: Blank1 Butyl acetate Specified Answer for: Blank2 Benzyl acetate Specified Answer for: Blank3 [None Given] Specified Answer for: Blank4 [None Given] Specified Answer for: Blank5 None Given]

Explanation / Answer

Question 1.

Answers :

HCOOH + CH3(CH2)2CH2OH -----> HCOOCH2CH2CH2CH3 [does not match with any you have selected here]

2, 3, and 4th are correct

Last,

CH3COOH + CH3(CH2)2OH -----> B. CH3COOCH2CH2CH3

Question 2.

1. Butyl formate

2. Phenylacetate

3. methylbutanoate

4. Phenylpropanoate

5. propylacetate

Question 3. If 2.03 g of salicyclic acid reacts with 61 mL of MeOH

reaction is 1: 1 molar ratio

calculate moles of salicyclic acid = g/molar mass = 2.03 g / 138.121 g/mol = 0.0147 mols

convert volume of methanol to mass = density x volume = 48.251 g

moles of MeOH = 1.506 mols

thus, salicyclic acid is the limiting reagent in this case

Calculate mass of product formed = moles of limiting reagent x molar mass of product ester

                                                       = 0.0147 x 152.1494 = 2.237 g

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