Question

For the reaction shown, compute the theoretical yield of the product in grams for each of the following initial amounts of reactants. 2Al(s)+3Cl2(g)→2AlCl3(s) 1.) 5.5 gAl; 19.8 gCl2 2.) 0.439 gAl; 2.29 gCl2

Answer 1

Atomi weight of Al =27 Cl =35.5

2Al(s)+3Cl₂(g)→2AlCl₃(s)

2 moles of Al reacts with 3 moles of Cl₂ to give 2 moles of AlCl₃

(2x27=54) grams of Al reacts with (3x71=213) grams of Cl₂ to give (2x133.5=267) grams of AlCl₃

1.) 5.5 gAl; 19.8 g Cl₂

    Here the limiting reagent is Cl₂ which has proportionally less mass compared to Al which has excess mass.

    So, the theoretical yield depends on mass of Cl₂

    213 grams of Cl₂ produce 267 grams of AlCl₃

    19.8 grams of Cl₂ produce ---? grams of AlCl₃   = (19.8/213)x267 = 24.82 grams

2.) 0.439 gAl; 2.29 gCl₂

    Here the limiting reagent is Al

    54 grams of Al produce 267 grams of AlCl₃

     0.439 grams of Al produce ----? grams of AlCl₃ = (0.439/54)x267 = 2.17 grams