If 34.0g Al and 39.0g Cl, how many g of AlCl3 are formed?

Question

The balanced equation is 2Al + 3Cl₂ ---> 2AlCl₃.

What would the yield be and how many grams of the excess reactant are left over? Please show work

Answer 1

In order to solve problems like these, we need to figure out which reagent is limiting. There are a few different ways to approach this, but one common approach is to calculate the amount of AlCl3 formed from assuming that each reagent is limiting.

For 34.0 g of Al we get:

34.0 g Al • 1 mol Al/ 26.98 g Al • 2 mol AlCl3/ 2 mol Al • 133.34 g AlCl3/ 1 mol AlCl3 = 168 g AlCl3

For 39.0 g of Cl₂ we get:

39.0 g Cl₂ • 1 mol Cl₂/ 70.9 g Cl₂ • 2 mol AlCl3/ 3 mol Cl2 • 133.34 g AlCl3 / 1 mol AlCl3 = 48.9 g AlCl3

Since 48.9 g AlCl3 is the smaller number this is the answer, and it also lets you know that Cl2 is the limiting reagent. We know this because once all the Cl2 is consumed we product 48.9 AlCl3 and will not have enough Cl2 left over to make anymore to get to 168 g.

Now we need to figure out how much excess aluminum is left over

39.0 g Cl₂ • 1 mol Cl₂/ 70.9 g Cl_{2 •} 2 mol Al/ 3 mol Cl2 • 26.98 g Al/ 1 mol Al = 9.89 g Al

34.0 g Al - 9.89 g Al = 24.1 g Al left over

Let me know if anything doesn't make sense!