Question

What volume of 0.140 HCl is needed to neutralize 2.58 of Mg(OH)2 ? ...?

Answer 1

To make a first step you have to know the balanced form for neutralization formula: $Mg(OH)2(aq)+2HCl(aq)--\ \textgreater \ 2H2O+Mg(aq)+Cl(aq)$
According to this, you can calculate what you are being asked :$0.0442molMg(OH)2(x)(2molHCl)/(1molMg(OH)2)=0.0885molHCl$
Then we have : $0.140MHCl=0.140(mol/L)HCl$
Hope everything is clear, here is the exact answer you need : $V=(0.0885molHCl)/(0.140(mol/L)HCl)=0.632LHCl$