Question

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)

Answer 1

1.65×10^-3 L or 1.65 mL

Step-by-step 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