What volume of 0.585 m ca(oh)2 would be needed to neutralize 15.8 l of 1.51 m hcl?

When the balanced reaction equation is:

2HCl(aq) + Ca(OH)2(aq) → CaCl2(aq) + 2H2O(l)

from the balanced equation, we can get the molar ratio between HCl & Ca(OH)2

2:1

∴ the volume of Ca(OH)2 = 15.8 L HCl * 1.51 m HCl * (1mol Ca(OH)2/ 2mol HCl) *                                           (1L ca(OH)2/0.585 mol Ca(OH)2 

                                          = 20.4 L

2.00 is the !