Question

A solution contains an unknown amount of dissolved magnesium. Addition of

Answer 1

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?

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

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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