moles of [tex]Ca(OH) _{2} [/tex] = 6.5 mol
mole ratio of [tex]Ca(OH) _{2} [/tex] : HCl = 1 : 2
∴ since [tex]Ca(OH) _{2} [/tex] cntain 6.5 mol, then HCL contains (2 * 6.5) mol
Thus moles of HCl = 13.0 mol
Volume of HCl needed = [tex] \frac{moles}{molar concentration} [/tex]
= [tex] \frac{13.0 mol}{1.5 mol / dm^{3} } [/tex]
= 8.67 [tex] dm^{3} [/tex]
[tex]since dm^{3} = L[/tex]
then volume of HCL needed = 8.67 L
