A student in an organic chemistry lab is told to prepare C2H5Br by the reaction of:...
- A student in an organic chemistry lab is told to prepare C2H5Br by the reaction of:
C2H5OH(l) + PBr3(aq) → C2H5Br(l) + H3PO3(aq)
- What is the limiting reagent?
- What is the theoretical yield of C2H5Br?
- What is the mass of excess reagent remaining?
- IF the student actually obtains 36.0g of C2H5Br, what is the percent yield?
Homework Answers
balanced reaction is
PBr3 + 3C2H5OH → 3C2H5Br + H3PO3
a. since 1 mol of PBr3 react with 3 mole of C2H5OH hence hence PBr3 is the limiting reagent
b. one mol of PBr3 give 3 mole of C2H5Br
mass = moles * Molar mass = 3 * 108.97 = 326.91 g
c. moles of excess reagent 3mol -1 mol = 2 mol
mass of excess reagent = 2 * 46 = 92 g
d. Perccent yield = (observed mass / theoretical mass ) * 100
= ( 36g / 326.91 ) *100 = 11%
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