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What volume of 0.126 M HCl is needed to neutralize 2.96g of Mg(OH)2?
The reaction equation is Mg(OH)2 + 2HCl = MgCl2 + H2O. The mol number of Mg(OH)2 is 2.96/58=0.051 mol. So we need 0.051*2=0.102 mol HCl to neutralize. The volume is 0.102/0.126=0.810 L =810 mL.
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