grapj9aylarinste grapj9aylarinste
  • 21-11-2016
  • Chemistry
contestada

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

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antonsandiego
antonsandiego antonsandiego
  • 22-11-2016
To make a first step you have to know the balanced form for neutralization formula: [tex]Mg(OH)2(aq)+2HCl(aq)--\ \textgreater \ 2H2O+Mg(aq)+Cl(aq) [/tex]
According to this, you can calculate what you are being asked :[tex]0.0442molMg(OH)2(x)(2molHCl)/(1molMg(OH)2)=0.0885molHCl[/tex]
Then we have : [tex]0.140MHCl=0.140(mol/L)HCl[/tex]
Hope everything is clear, here is the exact answer you need : [tex]V=(0.0885molHCl)/(0.140(mol/L)HCl)=0.632LHCl [/tex]

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