A solution contains an unknown amount of dissolved magnesium. Addition of
0.0877 mol of Na2CO3 causes complete precipitation of all of the magnesium.
What mass of magnesium was dissolved in the solution?

Respuesta :

Taking into account the reaction stoichiometry, 2.13 grams of magnesium was dissolved in the solution.

Reaction stoichiometry

In first place, the balanced reaction is:

Mg²⁺(aq) + Na₂CO₃(aq) → MgCO₃(s) + 2 Na⁺(aq)

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

  • Mg²⁺: 1 mole
  • Na₂CO₃: 1  mole
  • MgCO₃: 1 mole
  • Na⁺: 2 moles

The molar mass of the compounds is:

  • Mg²⁺: 24.3 g/mole
  • Na₂CO₃: 106 g/mole
  • MgCO₃: 84.3 g/mole
  • Na⁺: 23 g/mole

Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

  • Mg²⁺: 1 mole ×24.3 g/mole= 24.3 grams
  • Na₂CO₃: 1 mole ×106 g/mole= 106 grams
  • MgCO₃: 1 mole ×84.3 g/mole=84.3 grams
  • Na⁺: 2 moles ×23 g/mole= 46 grams

Mass of magnesium dissolved

The following rule of three can be applied: If by reaction stoichiometry 1 mole of Na₂CO₃ react with 24.3 grams of magnesium, 0.0877 moles of Na₂CO₃ react with how much mass of magnesium?

[tex]mass of magnesium=\frac{0.0877 moles of Na_{2}C O_{3}x24.3 grams of magnesium }{1 mole of Na_{2}C O_{3}}[/tex]

mass of magnesium= 2.13 grams

Finally, 2.13 grams of magnesium was dissolved in the solution.

Learn more about the reaction stoichiometry:

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