Question

exam 3 fall 2025
your initials nt

1. for the following mixtures/combination, if there is a reaction then write the products with their correct states, and balance the chemical equation. if there is no reaction then write nr.

(i) (5 points) hcl(aq)+ba(oh)₂(aq)→
(ii) (5 points) c₄h₁₀(g)+o₂(g)→
(iii) (5 points) li₂so₄(aq)+nh₄no₃(aq)→
(iv) (5 points) zn²⁺(aq)+al(s)→

1. (10 points (bonus)) balance the following redox reaction in acidic solution, using the half - reaction method.

fe²⁺(aq)+mno₄⁻(aq)→fe³⁺(aq)+mn²⁺(aq)

Explanation:

(I)

Step1: Determine the reaction type

This is an acid - base neutralization reaction. \(HCl\) is an acid and \(Ba(OH)_2\) is a base. The general form of an acid - base reaction is \(acid+base ightarrow salt + water\).

Step2: Write the unbalanced equation

\(HCl(aq)+Ba(OH)_2(aq) ightarrow BaCl_2(aq)+H_2O(l)\)

Step3: Balance the equation

Count the number of each atom. For \(Cl\) atoms: there is \(1\) \(Cl\) in \(HCl\) and \(2\) \(Cl\) in \(BaCl_2\), so we put a coefficient of \(2\) in front of \(HCl\). For \(H\) and \(O\) atoms: after putting \(2\) in front of \(HCl\), we have \(2H\) from \(HCl\) and \(2H\) and \(2O\) from \(Ba(OH)_2\). The balanced equation is \(2HCl(aq)+Ba(OH)_2(aq)=BaCl_2(aq)+2H_2O(l)\)

(II)

Step1: Determine the reaction type

This is a combustion reaction. The general form of a hydrocarbon combustion reaction is \(C_xH_y + O_2 ightarrow CO_2+H_2O\)

Step2: Write the unbalanced equation

\(C_4H_{10}(g)+O_2(g) ightarrow CO_2(g)+H_2O(l)\)

Step3: Balance the equation

First, balance the \(C\) atoms: put a coefficient of \(4\) in front of \(CO_2\). Then balance the \(H\) atoms: put a coefficient of \(5\) in front of \(H_2O\). Now, for the \(O\) atoms: we have \(4\times2 + 5\times1=13\) \(O\) atoms on the product side. So we put a coefficient of \(\frac{13}{2}\) in front of \(O_2\). To get rid of the fraction, we multiply the entire equation by \(2\). The balanced equation is \(2C_4H_{10}(g)+13O_2(g)=8CO_2(g)+10H_2O(l)\)

(III)

Step1: Check for a reaction

For a double - displacement reaction \(Li_2SO_4(aq)+NH_4NO_3(aq)\), we check the solubility rules. The possible products are \(LiNO_3\) and \((NH_4)_2SO_4\). Both \(LiNO_3\) and \((NH_4)_2SO_4\) are soluble in water. Since there is no precipitate, gas or weak electrolyte formed, there is no reaction. So the answer is \(NR\)

(IV)

Step1: Determine the reaction type

This is a single - displacement reaction. We use the activity series. Aluminum (\(Al\)) is more active than zinc (\(Zn\)).

Step2: Write the unbalanced equation

\(Zn^{2 +}(aq)+Al(s) ightarrow Al^{3+}(aq)+Zn(s)\)

Step3: Balance the equation

Use the electron - transfer method. The oxidation state of \(Al\) changes from \(0\) to \(+ 3\) (loses \(3e^-\)) and the oxidation state of \(Zn\) changes from \(+2\) to \(0\) (gains \(2e^-\)). The least common multiple of \(2\) and \(3\) is \(6\). So we put a coefficient of \(2\) in front of \(Al\) and \(3\) in front of \(Zn^{2+}\). The balanced equation is \(3Zn^{2+}(aq)+2Al(s)=2Al^{3+}(aq)+3Zn(s)\)

(6)

Step1: Write the half - reactions

Oxidation half - reaction: \(Fe^{2+}(aq) ightarrow Fe^{3+}(aq)\)
Reduction half - reaction: \(MnO_4^-(aq) ightarrow Mn^{2+}(aq)\)

Step2: Balance atoms other than \(O\) and \(H\) in each half - reaction

Oxidation: \(Fe^{2+}(aq) ightarrow Fe^{3+}(aq)\) (already balanced for \(Fe\))
Reduction: \(MnO_4^-(aq) ightarrow Mn^{2+}(aq)\) (already balanced for \(Mn\))

Step3: Balance \(O\) atoms in the reduction half - reaction

Add \(4H_2O\) to the right side: \(MnO_4^-(aq) ightarrow Mn^{2+}(aq)+4H_2O(l)\)

Step4: Balance \(H\) atoms in the reduction half - reaction (since it's in acidic solution)

Add \(8H^+\) to the left side: \(MnO_4^-(aq)+8H^+(aq) ightarrow Mn^{2+}(aq)+4H_2O(l)\)

Step5: Balance the charge in each half - reaction

Oxidation: \(Fe^{2+}(aq) ightarrow Fe^{3+}(aq)+e^-\)
Reduction: \(MnO_4^-(aq)+8H^+(aq)+5e^- ightarrow Mn^{2+}(aq)+4H_2O(l)\)

Step6: Multiply the half - reactions to make the number of electrons equal

Multiply the oxidation half - reaction by \(5\): \(5Fe^{2+}(aq) ightarrow 5Fe^{3+}(aq)+5e^-\)

Step7: Add the half - reactions together

\(5Fe^{2+}(aq)+MnO_4^-(aq)+8H^+(aq)=5Fe^{3+}(aq)+Mn^{2+}(aq)+4H_2O(l)\)

Answer:

(I) \(2HCl(aq)+Ba(OH)_2(aq)=BaCl_2(aq)+2H_2O(l)\)
(II) \(2C_4H_{10}(g)+13O_2(g)=8CO_2(g)+10H_2O(l)\)
(III) \(NR\)
(IV) \(3Zn^{2+}(aq)+2Al(s)=2Al^{3+}(aq)+3Zn(s)\)
(6) \(5Fe^{2+}(aq)+MnO_4^-(aq)+8H^+(aq)=5Fe^{3+}(aq)+Mn^{2+}(aq)+4H_2O(l)\)