Calculate the mass of MgCO3 precipitated by mixing 10.0 mL of a 0.200 M Na2CO3 solution with 5.00...


Calculate the mass of {eq}MgCO_3 {/eq} precipitated by mixing 10.0 mL of a 0.200 M {eq}Na_2CO_3 {/eq} solution with 5.00 mL of a 0.0400 M {eq}Mg(NO_3)_2 {/eq} solution.

Precipitation Reactions:

Precipitation reactions are reactions that result in the formation of a salt that does not ionize in the solutions after the reaction between two soluble compounds. In precipitation reactions, the mass of the precipitate formed is determined by the moles of the limiting reactant.

Answer and Explanation: 1

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We are given that:

  • {eq}[Na_2CO_3] = 0.200 \:M {/eq}
  • The volume of {eq}Na_2CO_3 {/eq} = 10.0 mL = 0.01 L
  • {eq}[Mg(NO_3)_2] = 0.04 \:M {/eq}
  • The...

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Learn more about this topic:

Precipitation Reactions: Predicting Precipitates and Net Ionic Equations


Chapter 10 / Lesson 9

Study precipitate reactions. Determine precipitate solubility according to solubility rules. Learn to write ionic equations by following a molecular reaction.

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