Suppose a student started with 133 mg of trans-cinnamic acid and 0520 mL of a 10

Question

Suppose a student started with 133 mg of trans-cinnamic acid and 0520 mL of a 10% (wi) bromine solution, and after the reaction and workup theoretical and percent yields. up with 0.205 g of brominated product. Calculate the student's Theoretical yield Percent yield Number Number 0.27 74.19 Incorrect OPrevous ®Give Up & View Solution # Try Agan ONextgExt Explanation The theoretical yield s the maximum gram amount of product that can be made assuming that the imiting reagent reacts to 100% completion. Did you use the moles of limiting reagent and molar mass of product to determine the theoretical yield? Did you use the equation below actual yield of product-×100 %=percent yield theoretical yield of product The density of the 10% bromine solution is 3.103 gmL The molar mass of bromine is 159.8 g/mo The molar mass of trans-cinnamic acid is 148 2 g/mol The molar mass of 2.3-bromo-3-phenylpropanoi acid product is 307.9 g/mol

Explanation / Answer

The involved reaction will be:

Cinnamic acid + Br2 - -- > 2,3-dibromo-3-phenylpropanoic acid

Here, Br2 volume = 0.520 mL x 0.1 = 0.052 mL

Br2 density = 3.102 g/mL

So, density = mass/ volume

Mass = density x volume

= 3.102 g/mL x 0.052

= 0.161 g

Moles of Br2 =mass / molar mass

= 0.161g / 159. 8 g mol-1

= 1.007 x 10^-3 mol

mw of trans-cinammic acid = 148. 2 g/mol

Moles of trans - cinnamic acid =0.133/148.2g

= 8.97 x 10^-4 mol

Since, mol of trans cinnamic acid present in lesser amount it is the limiting reagent.

So, theoretical yield = moles of trans-cinammic acid * molar mass of  2,3-dibromo-3- phenylpropanoic acid

=   8.97 x 10^-4 mol x 307.9g mol-1

= 0.276 g

So, percent yield = actual yield/ theoratical yield x 100

= 0.205/0.276 x 100

= 74.27%

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