How to find limiting reagent? i am given 4.3mL of isopentanol and acetic acid (4eq)
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Expert's answer
2018-10-01T08:02:02-0400
Isopentanol reacts with acetic acid to form ester. C5H11OH + CH3COOH = CH3COOC5H11 + H2O V(C5H11OH) = 4.3 mL density of isopentanol = 0.81 g/mL m(C5H11OH) = 0.81g/mL*4.3mL = 3.483 g M(C5H11OH) = 88 g/mol n(C5H11OH) = m/M = 3.483g/88g/mol = 0.04 mol n(CH3COOH) = 4 mol , because acetic acid is monobasic acid. According to the equation 1 mol of isopentanol reacts with 1 mol of acetic acid, so isopentanol is the limiting reagent, acetic acid is in the excess.
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