Question

1. What volume of 0.158 M HCl is required to neutralize 2.87 grams of Mg(OH)2?

2. Which ion(s) is/are spectator ions in the formation of a precipitate of BaSO4, when aqueous solutions of BaI2 and K2SO4 are mixed?

Answer 1

2 HCl + Mg(OH)2 --> MgCl2 + 2H2O

is the balanced reaction

given
mass of Mg(OH)2 = 2.87 g
molar mass of Mg(OH)2 = 58.32 g/mol

moles of Mg(OH)2 present = 1 mol/58.32g]* 2.87 g
= 0.049211 mols

from balanced equation it is clear that
mol ratio HCl: Mg(OH)2 = 2:1

mols of HCl needed = 2*0.049211 mols =0.098422 mols

molarity of HCl = 0.158 M

then volume needed = mols of HCl needed/molarity =0.098422 mols/0.158 M
=0.6229 l
round it for 3 significant figures

Answer Volume = 0.623 L
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In water both salts ionizes as

BaI2 (s) --> Ba2+ (aq) + 2I-(aq)

K2SO4 (s) ---> 2K+(aq) + SO42-(aq)

these ions combine to form product
complete ionic equation:

Ba2+ (aq) + 2I-(aq)+2K+(aq) + SO42-(aq) --> BaSO4(s) ++ 2I-(aq)+2K+(aq)

common ions present in product and reactant side are
called spectator ion
In this reaction spectator ions are I-(aq),K+(aq)
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cancel them to get

Net ionic equation

Ba2+(aq) + SO42-(aq) --> BaSO4(s)
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