Question

If 30.5 g of Al(OH)3 reacts with 39.8 g of HCl, how many grams of AlCl3 maximum could be produced?

Al(OH)3 + 3HCl ---> AlCl3 + H2O

Answer 1

To solve this problem, we need to find the limiting reagent. We can do this by converting both reactants to moles and comparing their mole ratios. The reactant that produces the smallest amount of product will be the limiting reagent.

30.5 g Al(OH)3 * (1 mol Al(OH)3/78 g Al(OH)3) = 0.39 mol Al(OH)3
39.8 g HCl * (1 mol HCl/36.5 g HCl) = 1.09 mol HCl

The mole ratio of Al(OH)3 to HCl is 1:3, so we need to multiply the moles of Al(OH)3 by 3 to compare the mole ratio.

0.39 mol Al(OH)3 * (3 mol HCl/1 mol Al(OH)3) = 1.17 mol HCl

Since we have more moles of HCl than Al(OH)3, Al(OH)3 is the limiting reagent.

Now we can calculate the maximum amount of AlCl3 that can be produced.

0.39 mol Al(OH)3 * (1 mol AlCl3/1 mol Al(OH)3) * (133.34 g AlCl3/1 mol AlCl3) = 52.2 g AlCl3

Therefore, the maximum amount of AlCl3 that can be produced is 52.2 g.