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