Given the balanced neutralization equation from part B, how many moles of potassium hydroxide (KOH) are required to neutralize 4.5 mol of sulfuric acid (H2SO4)? Assume that the sulfuric acid completely dissociates in water.

9 moles of KOH

Explanation:

Let's begin with the neutralization reaction.

Neutralization reaction always gives as product, water and a salt

2KOH + H₂SO₄ → 2H₂O + K₂SO₄

Ratio in the reactant side is 1:2. We can determine this rule of three:

1 mol of sulfuric acid can react with 2 moles of potassium hydroxide

Therefore, 4.5 moles sulfuric acod wil react with (4.5 . 2) /1 = 9 moles of KOH

the density of gold = 19.3 g/cm3

the relation between density and mass and volume is

density = mass / volume

given :

mass of gold = 13 g

density = 13g / volume = 19.3

volume = 13 /19.3 = 0.67 cm^3

thus volume of gold nugget = 0.67 cm^3