How many grams of magnesium metal will react completely with 8.3 liters of 5.5 M HCl? Show all of the work needed to solve this problem. Mg (s) + 2HCl (aq) → MgCl2 (aq) + H2 (g)

Ans: around 547 g of Mg will react completely with HCl

Given reaction:

Mg (s) + 2HCl (aq) → MgCl2 (aq) + H2 (g)

Volume of HCl, V = 8.3 L

Molarity of HCl, M = 5.5 M i. e. 5.5 moles/L

# Moles of HCl = M*V = 5.5 moles/L 8.3 L = 45.65 moles

Based on the reaction stoichiometry 1 mole of Mg reacts with 2 moles of HCl

Therefore moles of Mg that would react completely with 45.7 moles of HCl is:

= 45.65/2 = 22.825 moles

Atomic mass of Mg = 24 g/mole

Mass of Mg = 22.825 moles*24 g/mole = 547.8 g