Question

ILLUUSLIU15. 1) 2) Constants: A: MW = 150 g/mol; B: =MW 100 g/mol; C: MW = 200 g/mol. 2.0 g C was made from 4.5 g A and 4.0 g B Calculate the % yield for the following reaction: 3A + B C Calculate the % yield for the following reaction: 2A + 3B C Calculate the % yield for the following reaction: A+B 3C

explain how to calculate % yield.

Answer 1

Solution 3 A + B C Moles of A 0.03. ) 4.5 150 ) 4 = Moles of B 0.04 TOD Moles of c ) 2 ₂ 0.01 202 Molar ratio of A = 0.03 0.01 L 3 Molar ratio of B:7 0.04 0.04 So A is limiting reagent. Now, 3 moles of A & Imole of c. - 0.03 moles of A E L 003) o ol mole ofc Hence 1 yeilda 100= 0.01 1007) 100%. Actual Yeild Theoritical Yeild 001

2 A+ 3B 6 Molar ratio of A 47 0.03 0.015 Molay ratio of 6 = 0.04 e 0.0133 B is limiting reagent Now, 3 moles of B E Imole 0.04 moles of B E of x o 04 » OF 0.013. yeildo 100 Actual Yeild x Theoritical Yeild 00 x 100 = 0.76 X 1002 76 0.013 A + B 30 Molar ratio of A 22 0.03 0.03 Molar ratio of B2) 0.04 0.04 so A is limiting reagent. Imole of A = 3 moles ofc " 0.03 mole of Az 3 x 0.03 0.09 7. Yeilda Actual Heild Theoritical Yeild X100 ) 0.01 x 100 oog ( 0 11x 100 =)