Question

For the reaction shown, compute the theoretical yield of the product in grams for each of...

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)

Part B: 5.5 g Al; 19.8 g Cl2

Part C: 0.439 g Al; 2.29 g Cl2

Express your answers using three significant figures.

Thank you! :)

Homework Answers

Answer #1

2Al(s)+3Cl2(g)?2AlCl3(s)

So theoritically 2 mol of Al react with 3 mol of Cl2 to give 2 mol of AlCl3

=> 2 * 26.981 gm of Al + 3 * 70.906 gm of Cl2 ----------------> 2 * 133.34 gm of AlCl3

Part -B

5.5 gm of Al +19.8 gm of Cl2---> ?

5.5 gm of Al should react with = 3 * 70.906 / 2 * 26.981 *5.5 =21.681 gm of Cl2

So the limiting reagent is the Cl2

so 3 * 70.906 gm of Cl2 ----------> 2 * 133.34 gm of AlCl3

=> 19.8 gm of Cl2 will produce =  2 * 133.34 /  3 * 70.906 * 19.8 = 24.822 gm of AlCl3

Part C-

0.439 gm of Al + 2.29 gm of Cl2-------------------> ?

In the similar fashion 0.439 gm of Al will react with 1.730 gm og Cl2

So in this case Al is the limiting reagent

So 0.439 gm of Al will produce = 2 * 133.34 /  2*26.981 * 0.439 = 2.169 gm of AlCl3

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