Question

What mass of solid magnesium hydroxide is needed to neutralize .385 L of 2.5 M nitric acid?

Answer 1

2HNO3+Mg(OH)2>>Mg(NO3)2 + 2 H2O

so you need 1/2 mole magnesium hydroxide for each mole of nitric acid.

MoleNitricAcid=.385*2.5*(molmassNitricacid)

= .385*2.5*63

MassMg(OH)2=above/2 * molmassMgOH2
= above/2 * 58.3 grams

Answer 2

Mg(OH)2 + 2HNO3 => Mg(NO3)2 + 2HOH

moles = Molarity x Volume(L)
moles HNO3 = (2.5M)(0.385L) =(2.5mol/L)(0.385L)= 0.963mole HNO3

From balanced equation, half as many moles of Mg(OH)2 are needed as the amount of HNO3 given. That is, moles of Mg(OH)2 needed is 1/2(0.963 mole) = 0.481 mole Mg(OH)2 used to neutralize 0.963mole of HNO3. Converting moles to grams by multiplying by formula wt of Mg(OH)2 (= 58.33 g/mol) => (0.481mol)(58.33 g/mol) = 28.1 gms of Mg(OH)2 needed to neutralize .385L of 2.5M HNO3.