Question

For each of the following reactions, 13.0 g of each reactant is present initially. Determine the limiting reactant. 2Al(s)+3Cl2(g)→2AlCl3(s). Calculate the grams of product in parentheses that would be produced. (AlCl3)

Answer 1

Molar mass of Al = 26.981539 g/mol ;

We have 13 grams of Al and 13 grams of Cl2 as mentioned in the question.

So number of mkoles of Al in 13 grams = weight / molar mass

= 13 / 26.981539

= 0.4818 moles of Al are present in 13 grams.

Molar mass of Cl_{2 =} 70.9060 g/mol;

Number of moles of Cl₂ = 13 / 70.9060

= 0.1833 moles of Cl2 are present in 13 grams.

From the reaction equation given, we know that for two moles of Al, 3 moles of Cl2 are needed.

Therefore for 0.4818 moles of Al, 0.7227 moles of Cl2 will be needed but we have got only 0.1833 moles of Cl2.

So Cl2 is the limiting reagent.

0.1833 moles of Cl2 would react with only 0.1222 moles of Al.

Weight of 0.1222 moles of Al = 0.1222 x 26.981539 = 3.2971 grams

Molar mass of AlCl3 = 133.34 g/mol ;

If 26.981539 grams of Al can produce 133.34 grams of AlCl3

Then the amount of AlCl3 produced from 13 grams of Al would be = 64.244 grams ;

But actual reacting amount of Al is only 3.2971 grams

So the amount of AlCl3 produced from 3.2971 grams of Al = 16.2939 grams