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)

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! :)

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