how many mL of 1.8M HC2H3O2 are needed to neutralize 45mL of 1.4 M Al(OH)3

Question

 I need help solving this equation quick in under 3 minutes what is the best possible solution I can do

Answer 1

start by writing the balanced equation for the reaction. This is acid-base reaction.

 

    3 HC₂H₃O₂       +      Al(OH)₃   ___________   Al(C₂H₃O₂)₃      +   3 H₂O

 

Thus 3 mol HC2H3O2  is required to neutralize every 1 mol Al(OH)3.

 

The mol of Al(OH)3 neutralized = Molarity x Volume  =  1.4 mol/L   x 0.045 L = 0.063 mol.

 

Thus, 3 x 0.063 mol of HC2H3O2 is required.  The volume of HC2H3O2 required = mol / Molarity = 0.189 mol/(1.8 mol/L)   = 0.105 L = 105 ml.

 

105 ml of 1.8 M HC2H3O2 are needed.