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Chapter 4: Problem 53

Write balanced molecular and net ionic equations for the reactions of (a) manganese with dilute sulfuric acid, (b) chromium with hydrobromic acid, (c) tin with hydrochloric acid, (d) aluminum with formic acid, HCOOH.

Short Answer

Expert verified
\( (a) \) Molecular: \(Mn + H_2SO_4 \rightarrow MnSO_4 + H_2 \) Net ionic: \(Mn + 2H^+ \rightarrow Mn^{2+} + H_2 \) \( (b) \) Molecular: \(2Cr + 6HBr \rightarrow 2CrBr_3 + 3H_2 \) Net ionic: \(2Cr + 6H^+ \rightarrow 2Cr^{3+} + 3H_2 \) \( (c) \) Molecular: \(Sn + 2HCl \rightarrow SnCl_2 + H_2 \) Net ionic: \(Sn + 2H^+ \rightarrow Sn^{2+} + H_2 \) \( (d) \) Molecular: \(2Al + 6HCOOH \rightarrow 2Al(HCOO)_3 + 3H_2 \) Net ionic: \(2Al + 6HCOOH \rightarrow 2Al(HCOO)_3 + 3H_2 \)

Step by step solution

01

(a) Manganese with dilute sulfuric acid: Molecular equation

First, write the unbalanced molecular equation: Mn + H2SO4 -> MnSO4 + H2. Now, balance the equation by ensuring equal numbers of each atom on both sides of the equation. The balanced molecular equation is: Mn + H2SO4 -> MnSO4 + H2.
02

(a) Manganese with dilute sulfuric acid: Net ionic equation

Separate all strong electrolytes into their ions: Mn + 2H^+ + SO4^2- -> Mn^2+ + SO4^2- + H2. Now write the net ionic equation by removing the spectator ions: Mn + 2H^+ -> Mn^2+ + H2.
03

(b) Chromium with hydrobromic acid: Molecular equation

Write the unbalanced molecular equation: Cr + HBr -> CrBr3 + H2. Now, balance the equation by ensuring equal numbers of each atom on both sides of the equation. The balanced molecular equation is: 2Cr + 6HBr -> 2CrBr3 + 3H2.
04

(b) Chromium with hydrobromic acid: Net ionic equation

Separate all strong electrolytes into their ions: 2Cr + 6H^+ + 6Br^- -> 2Cr^3+ + 6Br^- + 3H2. Now write the net ionic equation by removing the spectator ions: 2Cr + 6H^+ -> 2Cr^3+ + 3H2.
05

(c) Tin with hydrochloric acid: Molecular equation

Write the unbalanced molecular equation: Sn + HCl -> SnCl2 + H2. Now, balance the equation by ensuring equal numbers of each atom on both sides of the equation. The balanced molecular equation is: Sn + 2HCl -> SnCl2 + H2.
06

(c) Tin with hydrochloric acid: Net ionic equation

Separate all strong electrolytes into their ions: Sn + 2H^+ + 2Cl^- -> Sn^2+ + 2Cl^- + H2. Now write the net ionic equation by removing the spectator ions: Sn + 2H^+ -> Sn^2+ + H2.
07

(d) Aluminum with formic acid: Molecular equation

Write the unbalanced molecular equation: Al + HCOOH -> Al(HCOO)3 + H2. Now, balance the equation by ensuring equal numbers of each atom on both sides of the equation. The balanced molecular equation is: 2Al + 6HCOOH -> 2Al(HCOO)3 + 3H2.
08

(d) Aluminum with formic acid: Net ionic equation

In this reaction, formic acid does not dissociate completely, so we do not separate it into its ions. The complete ionic equation is the same as the molecular equation: 2Al + 6HCOOH -> 2Al(HCOO)3 + 3H2. The balanced net ionic equation is: 2Al + 6HCOOH -> 2Al(HCOO)3 + 3H2.

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Most popular questions from this chapter

Specify what ions are present upon dissolving each of the following substances in water: (a) \(\mathrm{MgI}_{2}\), (b) \(\mathrm{K}_{2} \mathrm{CO}_{3}\), (c) \(\mathrm{HClO}_{4}\), (d) \(\mathrm{NaCH}_{3} \mathrm{COO}\).

A solution is made by mixing \(15.0 \mathrm{~g} \mathrm{of} \mathrm{} \mathrm{Sr}(\mathrm{OH})_{2}\) and \(55.0 \mathrm{~mL}\) of \(0.200 \mathrm{M} \mathrm{HNO}_{3}\). (a) Write a balanced equation for the reaction that occurs between the solutes. (b) Calculate the concentration of each ion remaining in solution. (c) Is the resultant solution acidic or basic?

Hard water contains \(\mathrm{Ca}^{2+}, \mathrm{Mg}^{2+}\), and \(\mathrm{Fe}^{2+}\), which interfere with the action of soap and leave an insoluble coating on the insides of containers and pipes when heated. Water softeners replace these ions with \(\mathrm{Na}^{+}\). Keep in mind that charge balance must be maintained. (a) If \(1500 \mathrm{~L}\) of hard water contains \(0.020 \mathrm{M} \mathrm{Ca}^{2+}\) and \(0.0040 \mathrm{M} \mathrm{Mg}^{2+}\), how many moles of \(\mathrm{Na}^{+}\) is needed to replace these ions? (b) If the sodium is added to the water softener in the form of \(\mathrm{NaCl}\), how many grams of sodium chloride are needed?

Identify the precipitate (if any) that forms when the following solutions are mixed, and write a balanced equation for each reaction. (a) \(\mathrm{NaCH}_{3} \mathrm{COO}\) and \(\mathrm{HCl}\), (b) \(\mathrm{KOH}\) and \(\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}\), (c) \(\mathrm{Na}_{2} \mathrm{~S}\) and \(\mathrm{CdSO}_{4}\).

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 \(0.0105 \mathrm{M} \mathrm{HCl}\). The \(\mathrm{NH}_{3}\) reacts with \(\mathrm{HCl}\) 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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