Question

Determine the number of grams of HCl that can react with 0.750 g Al(OH)3 according to the following reaction Al(OH)₃(s) + 3HCl(aq) → AlCl₃(aq) + 3H₂O(aq)

Answer 1

Answer:

1.053 grams of HCl.

Explanation:

1st) It is necessary to make sure that the equation is balanced:

$Al(OH)_3+3HCl\rightarrow AlCl_3+3H_2O$

From the balanced equation we know that 1 mole of Al(OH)3 react with 3 moles of HCl.

2nd) With the stoichiometry of the reaction and the molar mass of Al(OH)3 (78g/mol) and HCl (36.5g/mol) we can calculate the grams of HCl that can react with 0.750g:

$0.750g\text{ Al\lparen OH\rparen}_3*\frac{1mol\text{ Al\lparen OH\rparen}_3\text{ }}{78g\text{ Al\lparen OH\rparen}_3}*\frac{3mol\text{ HCl}}{1mol\text{ Al\lparen OH\rparen}_3}*\frac{36.5g\text{ HCl}}{1mol\text{ HCl}}=1.053g\text{ HCl}$

So, 1.053 grams of HCl can react with 0.750g of Al(OH)3.