Answer to Question #86535 in General Chemistry for Cristopher Gonzaga

Question

1.87 g acetic acid reacts with 2.13 g of isopentyl alcohol to yield isopentyl acetate. Find LR and ER.

Answer 1

1.     1.87 g acetic acid reacts with 2.13 g of isopentyl alcohol to yield isopentyl acetate. Find LR and ER.

m (CH3COOH) = 1.87 g

m (C5H12O) = 2.13 g

M (CH3COOH) = 60 g/mol

M (C5H12O) = 88 g/mol

M (Isopentyl acetate) = 130 g/mol

Limiting reactant – ?

Excess reactant – ?

Solution :

Step 1: Balance equation:

Step 2 and Step 3: Converting mass to moles and stoichiometry

n (CH3COOH) = m (CH3COOH) / M (CH3COOH) = 1.87 g / 60 g/mol = 0.031 mol

n (CH3COOH) = n (Isopentyl acetate) = 0.031 mol

m1 (Isopentyl acetate) = n (Isopentyl acetate)* M (Isopentyl acetate) = 0.031 mol*130 g/mol = 4.03 g

n (C5H12O) = m (C5H12O) / M (C5H12O) = 2.13 g / 88 g/mol = 0.024 mol

n (C5H12O) = n (Isopentyl acetate) = 0.024 mol

m2 (Isopentyl acetate) = 0.024 mol*130 g/mol = 3.12 g

Answer:The reactant that produces a smaller amount of product is the limiting reagent

isopentyl alcohol produces less Isopentyl acetate than does acetic acid (3.12 g Isopentyl acetate vs. 4.03 g Isopentyl acetate), therefore isopentyl alcohol is the limiting reagent in this reaction.

 The reactant that produces a larger amount of product is the excess reagent

acetic acid produces more amount of Isopentyl acetate than isopentyl alcohol (4.03 g Isopentyl acetate vs. 3.12 g Isopentyl acetate), therefore acetic acid  is the excess reagent in this reaction.