Question

question 2
10 pts
a precipitation reaction occurs when potassium carbonate is reacted with magnesium chloride. what is the balanced total molecular equation for this reaction.
○ k₂co₃ (aq) + mgcl (aq) → mgco₃ (s) + k₂cl (aq)
○ k₂co₃ (aq) + mgcl₂ (aq) → mgco₃ (aq) + 2kcl (s)
○ kco₃ (aq) + mgcl₂ (aq) → mgco₃ (s) + kcl₂ (aq)
○ k₂co₃ (aq) + mgcl₂ (aq) → mgco₃ (s) + 2kcl (aq)
○ kco₃ (aq) + mgcl₂ (aq) → mgco₃ (aq) + kcl₂ (s)
○ k₂co₃ (aq) + mgcl (aq) → mgco₃ (aq) + k₂cl (s)

Explanation:

Step1: Write reactants and products

Potassium carbonate ($K_2CO_3$) reacts with magnesium chloride ($MgCl_2$). The products of the double - displacement reaction are magnesium carbonate ($MgCO_3$) and potassium chloride ($KCl$). The general form of a double - displacement reaction is $AB + CD
ightarrow AD+CB$.

Step2: Balance the equation

Count the number of each type of atom on both sides of the equation. On the left - hand side, we have 2 potassium (K) atoms, 1 carbon (C) atom, 3 oxygen (O) atoms, 1 magnesium (Mg) atom, and 2 chlorine (Cl) atoms. On the right - hand side, to balance the K and Cl atoms, we need 2 moles of $KCl$. The balanced molecular equation is $K_2CO_3(aq)+MgCl_2(aq)
ightarrow MgCO_3(s) + 2KCl(aq)$. Also, according to solubility rules, $MgCO_3$ is insoluble and forms a precipitate (s), while $KCl$ is soluble (aq).

Answer:

$K_2CO_3(aq)+MgCl_2(aq)
ightarrow MgCO_3(s) + 2KCl(aq)$