Question

question 3
1 pts
what type of reaction is the following: ( mathrm{pb}(mathrm{no}_{3})_{2}(mathrm{aq})+mathrm{ki}(mathrm{aq})
ightarrow ) ?
single replacement
double replacement
decomposition
synthesis
question 4
1 pts
predict the products for the following reaction (it must be balanced): ( mathrm{pb}(mathrm{no}_{3})_{2}(mathrm{aq})+mathrm{ki}(mathrm{aq})
ightarrow ) ? + ?
( mathrm{kno}_{3}(mathrm{aq})+mathrm{pbl}_{2}(mathrm{~s}) )
( 2 mathrm{kno}_{3}(mathrm{aq})+mathrm{pbl}_{2}(mathrm{aq}) )
( 2 mathrm{kno}_{3}(mathrm{aq})+mathrm{pbl}_{2}(mathrm{~s}) )
( 2 mathrm{kno}_{3}(mathrm{~s})+mathrm{pbl}_{2}(mathrm{aq}) )

Explanation:

Question 3

Step1: Recall reaction types

Single - replacement: \(A + BC
ightarrow AC + B\). Double - replacement: \(AB+CD
ightarrow AD + CB\). Decomposition: \(AB
ightarrow A + B\). Synthesis: \(A + B
ightarrow AB\).

Step2: Analyze given reaction

For \(Pb(NO_3)_2(aq)+KI(aq)\), it follows \(AB + CD\) (where \(A = Pb^{2 +}\), \(B=NO_3^{-}\), \(C = K^{+}\), \(D = I^{-}\)) and will form \(AD+CB\) (double - replacement).

Question 4

Step1: Write the unbalanced reaction

\(Pb(NO_3)_2(aq)+KI(aq)
ightarrow KNO_3(aq)+PbI_2(s)\) (by double - replacement rules, \(PbI_2\) is insoluble (s) as per solubility rules).

Step2: Balance the equation

For \(Pb\): 1 on both sides. For \(NO_3^{-}\): 2 on left, so put 2 in front of \(KNO_3\). For \(K^{+}\): 2 on right, so put 2 in front of \(KI\). For \(I^{-}\): 2 on right (from \(PbI_2\)) and 2 on left (from \(2KI\)). The balanced equation is \(Pb(NO_3)_2(aq)+2KI(aq)
ightarrow 2KNO_3(aq)+PbI_2(s)\)

Answer:

Question 3: B. Double Replacement
Question 4: C. \(2KNO_3\ (aq)+PbI_2\ (s)\)