Question

Need to change to grams 5. When 3.3 g of C Hio is combined with 0.50 mL of liquid bromine, 3.5 g of C H1oBrs is isolated. What is the percent yield for this reaction? The density of liquid bromine is 13.1 g/mL. Ans. 45% yield I mol Bre ?p=m 0.50 mL Brex pala CH10(0) + 3 Brz () - CH10Bro (0 3.59 3.39

Answer 1

Here, first, we have to identify the limiting reagent in the reaction as according to that, we can calculate the theoretical yield of the reaction. As shown below, we have calculated the Actual mole ratio of the reactants then the stoichiometric ratio of reactants and comparison of those gave us the information that bromine is the limiting reagent, as we only required 0.33 moles of C7H10 per mole of bromine but in reality, we had an excess amount i.e. 0.858 moles of C7H10. The rest of the calculation is shown below-

CHOW+ 38*28)— C4 Htio Balc (2) 2 = m B m² 13.1 9 x 0.5 mL mL . 6.ssa C55 Et -=0.0409 159.89 mol mol B12 Mole ratio of reactants: moles of CG HiO 3.3g moles of Br2 0.0409 c.0351 mol Cho = qug.mat = 0.0409 mol B92 mol 20-858 mol Cytteo mol B:12 Stoichiometric ratio of reactants I mol cq hio 0.33 mol CH 10 Ional Brz 3 mol Biz Hence, the limiting reagent is Biz Hence, Theoretic = Imol 0.0409 mol В-1, x Cto Зе Smol Bilz = 0.0136 mel Catro B 576 X 573'4 g mot 7.817 9 CzHIo B176 44.7 ~ 457. 3. 5G X 100 7.8179 = I x 100 z