Question

Hydrochloric acid reacts with solid aluminum hydroxide to form aluminum chloride and water. How many mL of (2.4300x10^-1) M hydrochloric acid are needed to react completely with (3.22x10^0) grams of aluminum hydroxide?

Answer 1

no of moles of Al(OH)3   = W/G.M.Wt

                                     = 3.22/78   = 0.0413moles

Hydrochloric acid reacts with solid aluminum hydroxide to form aluminum chloride and water.

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

1 mole of Al(OH)3 react with 3 moles of HCl

0.0413 moles of Al(OH)3 react with = 3*0.0413/1   = 0.1239moles of HCl

molarity   = no of moles/volume of solution in L

0.243    = 0.1239/volume of solution in L

volume of solution in L   = 0.1239/0.243    = 0.5098L

                                   = 509.8ml >>>>>answer