Question

Al(OH)3(s) + HCl(aq) --> AlCl3(aq) + H2O(I) a) balance the equation b)calculate the number of grams s of HCl that can reacct with 0.500 g of Al(OH)3. c) calculate the number of AlCl3 and the number of grams of H2O formed when 0.500g of Al(OH)3 reacts. d) show that your calculations from parts b and c are consistent with the law of conservation of mass.

Answer 1

Al(OH)3 (s) + HCl (aq) ----> AlCl3(aq) + H2O(l)

The balanced eqaution is

a) Al(OH)3(s) + 3HCl(aq) --> AlCl3(aq) + 3H2O(I)

b) Molecular weights

Al(OH)3 78 HCl 36.5

moles of Al(OH)3= 0.5/78=0.00641 moles

1mole of Al(OH)3 requires 3 moles of HCl ( This is as per Stoichiometry)

0.00641 moles require =0.00641*3= 0.019231moldes mass of HCl = 0.019231*36.5 =0.701923 gms

moles of AlCl3 formed = 0.00641 (same as that of Al(OH)3

Mass of AlCl3 formed= 0.00641*133.5(Molecular weight of AlCl3)= 0.855769gms

Mass of Water formed =0.00641*3*18 ( 3moles of water are formed)=0.346

law of conservation of mass states that mass can neither be created nor destroyed. It can be changed from one form to another.

The mass of HCl is arried at based on the stoichiometry which states that

78gms of Al(OH)3 and 3*36.5 gms of HCl reacts to form 133.5 gms of AlCl3 and 54 gms of water

78+ 109.5 = 187.5gms mass of reactant mans of products = 133.5+54= 187.5

law of conservation of mass is obeyed.