Question

double displacement reactions
a precipitation reaction occurs when two solutions are mixed and an insoluble product is formed.
solubility is the ability of a substance to dissolve in water and the solubility rules will allow you to
predict whether a substance is soluble. looking at the reactants given in each we can see that the
compounds are soluble by the (aq) written beside them. if a product is soluble you should indicate
that with the (aq) symbol. an insoluble product is called a precipitate and is denoted with the (s)
symbol representing solid. first predict the possible products of the following reactions ensuring
that the formulas are correct based on the charge of the cation and anion, then include state of
matter and finally balance using coefficients.

1. ( mathrm{li}_{2} mathrm{~s}_{(mathrm{aq})}+mathrm{agno}_{3(mathrm{aq})} \to )
2. ( mathrm{bacl}_{2(mathrm{aq})}+mathrm{na}_{2} mathrm{co}_{3(mathrm{aq})} \to )
3. ( mathrm{alcl}_{3(mathrm{aq})}+mathrm{na}_{3} mathrm{po}_{4(mathrm{aq})} \to )
4. ( mathrm{ba}left(mathrm{no}_{3}

ight)_{2(mathrm{aq})}+mathrm{h}_{2} mathrm{so}_{4(mathrm{aq})} \to )

1. ( mathrm{k}_{2} mathrm{so}_{4(mathrm{aq})}+mathrm{caf}_{2(mathrm{aq})} \to )
2. ( mathrm{kcl}_{(mathrm{aq})}+mathrm{bano} 3_{(mathrm{aq})} \to )
3. ( mathrm{k}_{3} mathrm{po}_{4(mathrm{aq})}+left(mathrm{nh}_{4}

ight) mathrm{i}_{(mathrm{aq})} \to )

1. ( mathrm{li}_{3}left(mathrm{po}_{4}

ight)_{(mathrm{aq})}+mathrm{bacl}_{2(mathrm{aq})} \to )

1. ( mathrm{albr}_{3(mathrm{aq})}+mathrm{k}_{3} mathrm{po}_{4(mathrm{aq})} \to )
2. ( mathrm{zn}left(mathrm{no}_{3}

ight)_{2(mathrm{aq})}+mathrm{ba}(mathrm{oh})_{2(mathrm{aq})} \to )

Explanation:

Step1: Write the products

For \(Li_2S(aq)+AgNO_3(aq)\), the products are \(LiNO_3\) and \(Ag_2S\) based on double - displacement (cation - anion exchange).

Step2: Determine solubility

According to solubility rules:

• Nitrates (\(NO_3^-\)) are generally soluble. So \(LiNO_3\) is soluble (\(aq\)).
• Sulfides (\(S^{2 -}\)) of most metals (except group - 1 metals like \(Li\)) are insoluble. So \(Ag_2S\) is insoluble (\(s\)).

Step3: Balance the equation

The unbalanced equation is \(Li_2S(aq)+AgNO_3(aq)\to LiNO_3(aq)+Ag_2S(s)\)

• For \(Li\): There are \(2\) \(Li\) atoms on the left. So we need \(2\) \(LiNO_3\) on the right.
• For \(Ag\): There are \(2\) \(Ag\) atoms on the right, so we need \(2\) \(AgNO_3\) on the left.

The balanced equation is \(Li_2S(aq)+2AgNO_3(aq)\to 2LiNO_3(aq)+Ag_2S(s)\)

Answer:

\(Li_2S(aq)+2AgNO_3(aq)\to 2LiNO_3(aq)+Ag_2S(s)\)