Question

complete the balanced molecular chemical equation for the reaction below.
if no reaction occurs, write nr after the reaction arrow. be sure to include the proper phases
for all species within the reaction.
nacl(aq) + hg₂(c₂h₃o₂)₂(aq) →

Explanation:

Step1: Predict the products

When \(NaCl(aq)\) reacts with \(Hg_{2}(C_{2}H_{3}O_{2})_{2}(aq)\), a double - displacement reaction occurs. The general form of a double - displacement reaction is \(AB + CD
ightarrow AD+CB\). Here, \(A = Na^{+}\), \(B=Cl^{-}\), \(C = Hg_{2}^{2 +}\), and \(D = C_{2}H_{3}O_{2}^{-}\). The products are \(NaC_{2}H_{3}O_{2}\) and \(Hg_{2}Cl_{2}\).

Step2: Determine the solubility of products

According to solubility rules:

• Sodium acetate (\(NaC_{2}H_{3}O_{2}\)) is soluble in water. Compounds containing \(Na^{+}\) are generally soluble.
• Mercury(I) chloride (\(Hg_{2}Cl_{2}\)) is insoluble in water. Chloride salts are generally soluble, but \(Hg_{2}Cl_{2}\) is an exception.

Step3: Write the unbalanced equation

\(NaCl(aq)+Hg_{2}(C_{2}H_{3}O_{2})_{2}(aq)
ightarrow NaC_{2}H_{3}O_{2}(aq)+Hg_{2}Cl_{2}(s)\)

Step4: Balance the equation

Count the number of atoms of each element on both sides:

• For \(Na\): 1 on the left and 1 on the right.
• For \(Hg_{2}\): 1 on the left and 1 on the right.
• For \(C_{2}H_{3}O_{2}\): 2 on the left and 1 on the right.
• For \(Cl\): 1 on the left and 2 on the right.

Multiply \(NaCl\) by 2 and \(NaC_{2}H_{3}O_{2}\) by 2 to balance the equation.

Answer:

\(2NaCl(aq)+Hg_{2}(C_{2}H_{3}O_{2})_{2}(aq)
ightarrow 2NaC_{2}H_{3}O_{2}(aq)+Hg_{2}Cl_{2}(s)\)