Ask Question
15 April, 19:09

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)

+2
Answers (1)
  1. 15 April, 20:31
    0
    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
Know the Answer?
Not Sure About the Answer?
Get an answer to your 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 ...” in 📙 Chemistry if there is no answer or all answers are wrong, use a search bar and try to find the answer among similar questions.
Search for Other Answers