Question

chem semester 1 final 2024
questions #30-31, a student performs a lab experiment of reacting agno3 (aq) and cacl2 (aq) by mixing the two colorless solutions together to form a white powder (precipitate) in a colorless liquid.
ag (aq) + cacl2(aq) →
what products would you expect to be formed?
○ agno3 + cacl2
○ caag + clno3
○ agcl + ca(no3)2
○ no3 + ca + agcl

Explanation:

Step1: Identify Reaction Type

This is a double - displacement reaction. In a double - displacement reaction, the general form is \(AB + CD
ightarrow AD+CB\). Here, \(A = Ag^+\), \(B = NO_3^-\), \(C = Ca^{2+}\), \(D = Cl^-\) from the reactants \(AgNO_3(aq)\) and \(CaCl_2(aq)\).

Step2: Apply Double - Displacement Rules

For the reaction \(2AgNO_3(aq)+CaCl_2(aq)
ightarrow\), we swap the cations and anions. The silver ion (\(Ag^+\)) combines with the chloride ion (\(Cl^-\)) and the calcium ion (\(Ca^{2+}\)) combines with the nitrate ion (\(NO_3^-\)).
The formula for the product from \(Ag^+\) and \(Cl^-\) is \(AgCl\) (since \(Ag^+\) has a charge of \( + 1\) and \(Cl^-\) has a charge of \(- 1\), so \(AgCl\) is neutral). The formula for the product from \(Ca^{2+}\) and \(NO_3^-\) is \(Ca(NO_3)_2\) (since \(Ca^{2+}\) has a charge of \(+2\) and \(NO_3^-\) has a charge of \(- 1\), so we need two \(NO_3^-\) ions to balance the charge of one \(Ca^{2+}\) ion). Also, \(AgCl\) is insoluble in water (a precipitate, the white powder), and \(Ca(NO_3)_2\) is soluble.

Answer:

\(AgCl + Ca(NO_3)_2\) (corresponding to the option "AgCl + Ca(NO₃)₂")