Let's say you added too much magnesium in the amount of 0.1594 g. How much 8.00 M HCl (in mL) would be required to decompose the excess Mg present?

Mg (s) + 2 HCl (aq) - -> MgCl2 (aq) + H2 (g)

1.65*10^-3 L or 1.65 mL

Explanation:

Number of moles of magnesium in the excess reagent = mass of excess reagent / molar mass of magnesium

Mass of excess magnesium = 0.1594 g

Molar mass of magnesium = 24 gmol-1

Number of moles of magnesium = 0.1594/24 = 6.6*10^-3 moles

From the reaction equation;

1 mole of magnesium reacted with 2 moles of HCl

6.6*10^-3 moles of magnesium will react with 6.6*10^-3 moles * 2 = 13.2*10^-3 moles of HCl

But we know that;

Number of moles of HCl = concentration of HCl * volume of HCl

Volume of HCl = number of moles of HCl / concentration of HCl

Since concentration of HCl = 8.00M

Volume of HCl = 13.2*10^-3/8.00 = 1.65*10^-3 L or 1.65 mL