Question

1. According to the following reaction, how many grams of AlCl₃ will be formed if the reaction was started with 123 g of Cl₂ and excess Al?

2 Al + 3 Cl₂ → 2 AlCl₃

b). From the prior question, if 126.8 g AlCl₃ is actually recovered, what is the percent yield?

Answer 1

TS 2A + 3, 2AlCl3 Molar Mass : 70.1 g/mol 133.34 g/mol from well-balanced Equation, we know- 3 moles of l produces 2 moles of All, ie. (3x70.1) g of a produces (2 x 133.34) g of Alcea . 123 g of Cl₂ will produce (3x70.1) g 7 (2 x 133.34) g x 123g of All, (155.98 g ok At 42 Theoritical yield (6) Acc to Question - observed yield = 126.8 g observed yield x 100 7. Yield = Theoritical yield = 126.8 x 100 155.98 = (81.29.7.) Please Rate...