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Chapter 4: Problem 54

Write balanced molecular and net ionic equations for the reactions of (a) hydrochloric acid with nickel, (b) dilute sulfuric acid with iron, \((\mathbf{c})\) hydrobromic acid with magnesium, (d) acetic acid, \(\mathrm{CH}_{3} \mathrm{COOH},\) with zinc.

Short Answer

Expert verified
The balanced molecular and net ionic equations for the given reactions are: (a) Molecular: 2HCl + Ni -> NiCl₂ + H₂ Net ionic: 2H⁺ + Ni -> Ni⁺² + H₂ (b) Molecular: H₂SO₄ + Fe -> FeSO₄ + H₂ Net ionic: 2H⁺ + Fe -> Fe⁺² + H₂ (c) Molecular: 2HBr + Mg -> MgBr₂ + H₂ Net ionic: 2H⁺ + Mg -> Mg⁺² + H₂ (d) Molecular: 2CH₃COOH + Zn -> Zn(CH₃COO)₂ + H₂ Net ionic: 2CH₃COOH + Zn -> Zn⁺² + 2CH₃COO⁻ + H₂

Step by step solution

01

1. Write unbalanced molecular equations for the reactions

For each reaction, we need to write the molecular equation by writing down the reactants and products. We can do this using the given information: (a) Hydrochloric acid (HCl) with nickel (Ni) HCl + Ni -> NiCl₂ + H₂ (b) Dilute sulfuric acid (H₂SO₄) with iron (Fe) H₂SO₄ + Fe -> FeSO₄ + H₂ (c) Hydrobromic acid (HBr) with magnesium (Mg) HBr + Mg -> MgBr₂ + H₂ (d) Acetic acid (CH₃COOH) with zinc (Zn) CH₃COOH + Zn -> Zn(CH₃COO)₂ + H₂
02

2. Balance the molecular equations

Now that we have the molecular equations for each reaction, let's balance each equation. (a) 2HCl + Ni -> NiCl₂ + H₂ (b) H₂SO₄ + Fe -> FeSO₄ + H₂ (c) 2HBr + Mg -> MgBr₂ + H₂ (d) 2CH₃COOH + Zn -> Zn(CH₃COO)₂ + H₂
03

3. Write unbalanced net ionic equations

Back to our reactions, we can represent these reactions as net ionic equations, which show only the species that participate in the reaction and are changed during it. (a) 2H⁺ + 2Cl⁻ + Ni -> Ni⁺² + 2Cl⁻ + H₂ (b) 2H⁺ + SO₄⁻² + Fe -> Fe⁺² + SO₄⁻² + H₂ (c) 2H⁺ + 2Br⁻ + Mg -> Mg⁺² + 2Br⁻ + H₂ (d) 2CH₃COOH + Zn -> Zn⁺² + 2CH₃COO⁻ + H₂
04

4. Write balanced net ionic equations

Finally, remove the spectator ions (ions that are unchanged in the reaction) and balance the equations. (a) 2H⁺ + Ni -> Ni⁺² + H₂ (b) 2H⁺ + Fe -> Fe⁺² + H₂ (c) 2H⁺ + Mg -> Mg⁺² + H₂ (d) 2CH₃COOH + Zn -> Zn⁺² + 2CH₃COO⁻ + H₂ These are the balanced molecular and net ionic equations for the given reactions.

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Most popular questions from this chapter

Ritalin is the trade name of a drug, methylphenidate, used to treat attention- deficit/hyperactivity disorder in young adults. The chemical structure of methylphenidate is (a) Is Ritalin an acid or a base? An electrolyte or a nonelectrolyte? (b) A tablet contains a \(10.0-\mathrm{mg}\) dose of Ritalin. Assuming all the drug ends up in the bloodstream, and the average man has a total blood volume of \(5.0 \mathrm{~L}\), calculate the initial molarity of Ritalin in a man's bloodstream. (c) Ritalin has a half-life of 3 hours in the blood, which means that after 3 hours the concentration in the blood has decreased by half of its initial value. For the man in part (b), what is the concentration of Ritalin in his blood after 6 hours?

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Federal regulations set an upper limit of 50 parts per million (ppm) of \(\mathrm{NH}_{3}\) in the air in a work environment [that is, 50 molecules of \(\mathrm{NH}_{3}(g)\) for every million molecules in the air]. Air from a manufacturing operation was drawn through a solution containing $1.00 \times 10^{2} \mathrm{~mL}\( of \)0.0105 \mathrm{MHCl} .\( The \)\mathrm{NH}_{3}$ reacts with HCl according to: $$ \mathrm{NH}_{3}(a q)+\mathrm{HCl}(a q) \longrightarrow \mathrm{NH}_{4} \mathrm{Cl}(a q) $$ After drawing air through the acid solution for \(10.0 \mathrm{~min}\) at a rate of \(10.0 \mathrm{~L} / \mathrm{min},\) the acid was titrated. The remaining acid needed \(13.1 \mathrm{~mL}\) of \(0.0588 \mathrm{M} \mathrm{NaOH}\) to reach the equivalence point. (a) How many grams of \(\mathrm{NH}_{3}\) were drawn into the acid solution? (b) How many ppm of \(\mathrm{NH}_{3}\) were in the air? (Air has a density of \(1.20 \mathrm{~g} / \mathrm{L}\) and an average molar mass of \(29.0 \mathrm{~g} / \mathrm{mol}\) under the conditions of the experiment.) (c) Is this manufacturer in compliance with regulations?
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