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`