Verified Answer :

JoshEast

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

Related Questions