Find the volume of 0.120 M hydrochloric acid necessary to react completely with 1.51 g Al(OH)^3

Answer 1

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

moles Al(OH)3 = grams/molar mass
Using the coefficients in the balanced equation, convert moles Al(OH)3 to moles HCl.
Then M HCl = moles HCl/L HCl. You know M and moles HCl, solve for L.

Answer 2

.1613

Answer 3

0.4839L