Question

QUESTION 15 When 80.4 g Al and 70.4 g Cl2 react according to the following equation, how much AlCl3 can be made, assuming 100 percent yield? 2A1 + 3Cl2 + 2AlCl3 Hint: limiting reagent O 38.09 21.29 9.02 g 46.2 g 73.0 g 88.2 g 91.69 1959

Answer 1

Limiting reagent in this reaction is Cl2 . because mol of Cl2 / its coifficient is less than that of Al .

2 AL +30 2 Alcz Giver, mass of Al = 80.ug. cz - 70.4 g Mole of Al= mass of AL = 80.4 26.98 molecular mass = 2.9799 male Moles of l = 70.4 70.9 = 0.99 29 mal. Here, az -ration is is limiting reagent because Cl₂ concent- less. Al moles are 2.9799 which can be reacted B. with 3x2.9799 mole = 1.98 66 mole a moles will decide the product of Cly. so, formation. of a 3 mole 1 mole giveng gives " 2 mole of Alch 2 mole of Alce 2x0.9929 mol of Alce 0.9929. Mass of Alcez = mole x molecular wt. 2*0.9929 x 133:33 = 288.25 g - 88.2 g of Aldez is produced.