Question

exercise 3: chemical reactions in aqueous solutions
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1. predict the products of the following reactions. identify the precipitate and the spectator ions and write the net ionic equation.

chemical reactions precipitate spectator ions net ionic equation
a.) agno₃(aq) + nabr(aq) →
____ + ____
b.) k₃po₄(aq) + zn(ch₃coo)₂(aq) →
____ + ____
c.) na₂s(aq) + pb(no₃)₂(aq) →
____ + ____

1. complete and write the overall chemical equation, the complete ionic equation, and the net ionic equation for the reactions below. if the substance is weak acid or base, use its molecular form when writing the equations.

chemical reactions complete ionic equation net ionic equation
a.) koh(aq) + hcl(aq) →
b.) ch₃cooh(aq) + naoh (aq) →

1. give the oxidation number of the specified atom in the given molecules or ions.

molecules / ions oxidation numbers molecules / ions oxidation numbers
a.) xe in xef₄ d.) b in b₂o₃
b.) cl in hclo e.) as in h₃aso₃
c.) br in bro₄⁻ f.) v in vo²⁺

1. for each of the following equations, identify the substance oxidized, the substance reduced, the oxidizing agent, and the reducing agent.

chemical reactions substance oxidized substance reduced oxidizing agent reducing agent
1.) no(g) + o₃(g) → no₂(g) + o₂(g)
2.) 2 h₂s(g) + so₂(g) → 3 s(s) + 2 h₂o(l)

Explanation:

1a. Predict products, precipitate, spectator ions and net - ionic equation for \(AgNO_3(aq)+NaBr(aq)\)

Step1: Predict products

By double - displacement reaction, \(AgNO_3(aq)+NaBr(aq) ightarrow AgBr(s)+NaNO_3(aq)\). The precipitate is \(AgBr\) because silver bromide is insoluble in water. The spectator ions are \(Na^+\) and \(NO_3^-\) as they remain in solution unchanged.
The net ionic equation is \(Ag^+(aq)+Br^-(aq) ightarrow AgBr(s)\).

1b. Predict products, precipitate, spectator ions and net - ionic equation for \(K_3PO_4(aq)+Zn(CH_3COO)_2(aq)\)

Step1: Predict products

By double - displacement reaction, \(2K_3PO_4(aq)+3Zn(CH_3COO)_2(aq) ightarrow Zn_3(PO_4)_2(s)+6KCH_3COO(aq)\). The precipitate is \(Zn_3(PO_4)_2\) as it is insoluble. The spectator ions are \(K^+\) and \(CH_3COO^-\).
The net ionic equation is \(2PO_4^{3 -}(aq)+3Zn^{2+}(aq) ightarrow Zn_3(PO_4)_2(s)\).

1c. Predict products, precipitate, spectator ions and net - ionic equation for \(Na_2S(aq)+Pb(NO_3)_2(aq)\)

Step1: Predict products

By double - displacement reaction, \(Na_2S(aq)+Pb(NO_3)_2(aq) ightarrow PbS(s)+2NaNO_3(aq)\). The precipitate is \(PbS\) as it is insoluble. The spectator ions are \(Na^+\) and \(NO_3^-\).
The net ionic equation is \(Pb^{2+}(aq)+S^{2 -}(aq) ightarrow PbS(s)\).

2a. Complete and write equations for \(KOH(aq)+HCl(aq)\)

Step1: Overall chemical equation

\(KOH(aq)+HCl(aq) ightarrow KCl(aq)+H_2O(l)\)

Step2: Complete ionic equation

\(K^+(aq)+OH^-(aq)+H^+(aq)+Cl^-(aq) ightarrow K^+(aq)+Cl^-(aq)+H_2O(l)\)

Step3: Net ionic equation

\(H^+(aq)+OH^-(aq) ightarrow H_2O(l)\)

2b. Complete and write equations for \(CH_3COOH(aq)+NaOH(aq)\)

Step1: Overall chemical equation

\(CH_3COOH(aq)+NaOH(aq) ightarrow CH_3COONa(aq)+H_2O(l)\)

Step2: Complete ionic equation

\(CH_3COOH(aq)+Na^+(aq)+OH^-(aq) ightarrow CH_3COO^-(aq)+Na^+(aq)+H_2O(l)\)

Step3: Net ionic equation

\(CH_3COOH(aq)+OH^-(aq) ightarrow CH_3COO^-(aq)+H_2O(l)\)

3a. Oxidation number of \(Xe\) in \(XeF_4\)

Step1: Use rules

Let the oxidation number of \(Xe\) be \(x\). Fluorine has an oxidation number of \(- 1\). In \(XeF_4\), \(x + 4\times(-1)=0\), so \(x = + 4\).

3b. Oxidation number of \(Cl\) in \(HClO\)

Step1: Use rules

Hydrogen has an oxidation number of \(+1\) and oxygen has an oxidation number of \(-2\). Let the oxidation number of \(Cl\) be \(y\). Then \(+1 + y+(-2)=0\), so \(y = + 1\).

3c. Oxidation number of \(Br\) in \(BrO_4^-\)

Step1: Use rules

Oxygen has an oxidation number of \(-2\). Let the oxidation number of \(Br\) be \(z\). Then \(z+4\times(-2)= - 1\), so \(z = + 7\).

3d. Oxidation number of \(B\) in \(B_2O_3\)

Step1: Use rules

Oxygen has an oxidation number of \(-2\). Let the oxidation number of \(B\) be \(a\). Then \(2a+3\times(-2)=0\), so \(a = + 3\).

3e. Oxidation number of \(As\) in \(H_3AsO_3\)

Step1: Use rules

Hydrogen has an oxidation number of \(+1\) and oxygen has an oxidation number of \(-2\). Let the oxidation number of \(As\) be \(b\). Then \(3\times(+1)+b + 3\times(-2)=0\), so \(b = + 3\).

3f. Oxidation number of \(V\) in \(VO^{2+}\)

Step1: Use rules

Oxygen has an oxidation number of \(-2\). Let the oxidation number of \(V\) be \(c\). Then \(c+(-2)=+2\), so \(c = + 4\).

4a. Identify substances for \(NO(g)+O_3(g)

ightarrow NO_2(g)+O_2(g)\)

Step1: Oxidation states

In \(NO\), \(N\) has an oxidation number of \(+2\), in \(NO_2\), \(N\) has an oxidation number of \(+4\), so \(NO\) is oxidized. \(O_3\) is reduced as the oxygen in \(O_3\) has a higher oxidation state than in \(O_2\). The oxidizing agent is \(O_3\) and the reducing agent is \(NO\).

4b. Identify substances for \(2H_2S(g)+SO_2(g)

ightarrow 3S(s)+2H_2O(l)\)

Step1: Oxidation states

In \(H_2S\), \(S\) has an oxidation number of \(-2\), in \(SO_2\), \(S\) has an oxidation number of \(+4\), and in \(S(s)\) the oxidation number is \(0\). \(H_2S\) is oxidized and \(SO_2\) is reduced. The oxidizing agent is \(SO_2\) and the reducing agent is \(H_2S\).

Answer:

1a. Products: \(AgBr(s)+NaNO_3(aq)\), Precipitate: \(AgBr\), Spectator ions: \(Na^+\), \(NO_3^-\), Net ionic equation: \(Ag^+(aq)+Br^-(aq)
ightarrow AgBr(s)\)
1b. Products: \(Zn_3(PO_4)_2(s)+6KCH_3COO(aq)\), Precipitate: \(Zn_3(PO_4)_2\), Spectator ions: \(K^+\), \(CH_3COO^-\), Net ionic equation: \(2PO_4^{3 -}(aq)+3Zn^{2+}(aq)
ightarrow Zn_3(PO_4)_2(s)\)
1c. Products: \(PbS(s)+2NaNO_3(aq)\), Precipitate: \(PbS\), Spectator ions: \(Na^+\), \(NO_3^-\), Net ionic equation: \(Pb^{2+}(aq)+S^{2 -}(aq)
ightarrow PbS(s)\)
2a. Overall: \(KOH(aq)+HCl(aq)
ightarrow KCl(aq)+H_2O(l)\), Complete ionic: \(K^+(aq)+OH^-(aq)+H^+(aq)+Cl^-(aq)
ightarrow K^+(aq)+Cl^-(aq)+H_2O(l)\), Net ionic: \(H^+(aq)+OH^-(aq)
ightarrow H_2O(l)\)
2b. Overall: \(CH_3COOH(aq)+NaOH(aq)
ightarrow CH_3COONa(aq)+H_2O(l)\), Complete ionic: \(CH_3COOH(aq)+Na^+(aq)+OH^-(aq)
ightarrow CH_3COO^-(aq)+Na^+(aq)+H_2O(l)\), Net ionic: \(CH_3COOH(aq)+OH^-(aq)
ightarrow CH_3COO^-(aq)+H_2O(l)\)
3a. \(+4\)
3b. \(+1\)
3c. \(+7\)
3d. \(+3\)
3e. \(+3\)
3f. \(+4\)
4a. Substance oxidized: \(NO\), Substance reduced: \(O_3\), Oxidizing agent: \(O_3\), Reducing agent: \(NO\)
4b. Substance oxidized: \(H_2S\), Substance reduced: \(SO_2\), Oxidizing agent: \(SO_2\), Reducing agent: \(H_2S\)