Question

help with limiting reagents and percent yield needed

Name Physical Properties of Regents and Products Structure and Molar Melting Mass Formula PL ("C) (g/mL) (g/mol) Boiling PL. ("C) (1) cis-stilbene | Cay Hiz (2) trans-stilbene CuHız 186.24 2011 5-6 307 180.24 122-135 305-307 3 319.822 -8,0 60.052 1.05 16.6 (3) pyridinium tribromide Cg Ho Brgn CH₃COOH (4) Glacial Acetic Acid Quantities of Reagents Used and Products Formed - cis-Stilbene Structural Formula Volume Mass Moles Limiting (mL) Reagent Check ta (442 Ca Hi Bro N 0.303 |(3) 0.6 0.0017 0.0019 0.105 0.008 06.0 CH₃COOH Product racenic mixture 2.6 Percent Yield: Quantities of Reagents Used and Products Formed - trans-Stilbene Structural Formula 1 Volume Mass Moles Limiting (mL) Reagent Check 100.4 0.0022 0.8 " CG Hz 26 BrsN Trens Ci4 H12 A CH3COOH meso compand 8.0 0.was 0.133 0.0018 roduct 0.6

the reactions are the bromination of cis-stilbene and trans-stilbene. The reagents and the moles of each reagent are in my data, as is the moles of the products formed. The volume and mass of the reagents are given as well. I need to know how to find the percent yield and what the limiting reagents are for both the trans and cis stilbene brominations. The information is there.

Answer 1

Solution:

The limiting reagent is the reagent whose number of moles is minimum. Thus, according to calculated number of moles, the number of moles of cis- stilbine in table 1and pyridimium tribromide in table 2 are minimum.

Thus,

1) Cis-stilbine is limiting reagent

Number of moles of cis stilbine = 0.0017 mol

Hence, theoretical number of moles of product = 0.0017 mol

Molar mass of product = 340.05 g /mol

Theoretical mass of product = number of moles x molar mass of product

= 0.0017 mol x 340.05 g/mol = 0.5780 g

Percent yield = (actual mass /theoretical mass) x 100

= (0.5780 g / 2.6 g) x 100 = 22.23 %

2) Pyridinium tribromide is the limiting reagent.

Number of moles of pyridinium tribromide = 0.0022 mol

Hence, theoretical moles of product = 0.0022 mol

Theoretical mass of product = number of moles x molar mass

= 0.0022 mol x 340.05 g/mol = 0.748 g

Percent yield = (0.6 g / 0.748g ) x100 = 80.21%