1-Chlorobutane (2.5 mL, d=0.886) in 20 mL of acetone is reacted with 90 mL of a

Question

1-Chlorobutane (2.5 mL, d=0.886) in 20 mL of acetone is reacted with 90 mL of a 15 wt% solution of NaI in acetone. After work-up, you obtain 1.3 g  of 1-iodobutane. which is the limiting reagent and what is your % yield?

Explanation / Answer

the reaction is 1:1. first calculate the number of mol of eachreactive: 1C-butane: 2.5ml --> 2.215 g   Pm = 93.52g/mol -->0.024mol NaI 90ml whit density 0.79g/cm3--> 71.1g dis. of this, the 15%are from NaI --> 10.665 g Pm = 149.89 g/mol --> 0.071 mol limiting reagent is Cl-butane. you must obtein the same amount ofI-butane 0.024 mol Pm = 185.02g/mol --> 4.44 g. if you obteined 1.3g, the yield is: R =29.3%

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