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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) Mn + 2H鈧係O鈧 鈫 MnSO鈧 + H鈧; Mn + 2H鈦 鈫 Mn虏鈦 + H鈧 (b) Cr + 2HBr 鈫 CrBr鈧 + H鈧; Cr + 2H鈦 鈫 Cr虏鈦 + H鈧 (c) Sn + 2HCl 鈫 SnCl鈧 + H鈧; Sn + 2H鈦 鈫 Sn虏鈦 + H鈧 (d) 2Al + 6HCOOH 鈫 2Al(HCOO)鈧 + 3H鈧; 2Al + 6H鈦 鈫 2Al(HCOO)鈧 + 3H鈧

Step by step solution

01

Write Molecular Equations

For each metal reacting with an acid, we need to write the balanced molecular equation. (a) Manganese with dilute sulfuric acid:\[ \text{Mn} + \text{H}_2\text{SO}_4 \rightarrow \text{MnSO}_4 + \text{H}_2 \](b) Chromium with hydrobromic acid:\[ \text{Cr} + 2\text{HBr} \rightarrow \text{CrBr}_2 + \text{H}_2 \](c) Tin with hydrochloric acid:\[ \text{Sn} + 2\text{HCl} \rightarrow \text{SnCl}_2 + \text{H}_2 \](d) Aluminum with formic acid, HCOOH:\[ 2\text{Al} + 6\text{HCOOH} \rightarrow 2\text{Al(HCOO)}_3 + 3\text{H}_2 \]
02

Dissociate the Ionic Compounds

Next, we need to dissociate any ionic compounds in aqueous solution into their ions. For simplicity, assume all salts are soluble and dissociate completely, and the acid is in aqueous solution.(a) Manganese sulfate:\[ \text{MnSO}_4 \rightarrow \text{Mn}^{2+} + \text{SO}_4^{2-} \](b) Chromium bromide:\[ \text{CrBr}_2 \rightarrow \text{Cr}^{2+} + 2\text{Br}^- \](c) Tin chloride:\[ \text{SnCl}_2 \rightarrow \text{Sn}^{2+} + 2\text{Cl}^- \](d) Aluminum formate does not dissociate easily and is kept as \( \text{Al(HCOO)}_3 \).
03

Identify Spectator Ions

Spectator ions are ions that appear the same on both sides of the equation. These ions do not participate in the overall reaction and will be removed in the net ionic equation. (a) There are no spectator ions in this reaction. (b) Again, there are no spectator ions. (c) No spectator ions. (d) No spectator ions, as we assume aluminum formate stays intact in the reaction.
04

Write Net Ionic Equations

Now we write the net ionic equations by removing the spectator ions and focusing on the chemical species that change.(a) Manganese with sulfuric acid:\[ \text{Mn} + 2\text{H}^+ \rightarrow \text{Mn}^{2+} + \text{H}_2 \](b) Chromium with hydrobromic acid:\[ \text{Cr} + 2\text{H}^+ \rightarrow \text{Cr}^{2+} + \text{H}_2 \](c) Tin with hydrochloric acid:\[ \text{Sn} + 2\text{H}^+ \rightarrow \text{Sn}^{2+} + \text{H}_2 \](d) Aluminum with formic acid:\[ 2\text{Al} + 6\text{H}^+ \rightarrow 2\text{Al(HCOO)}_3 + 3\text{H}_2 \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Molecular Equations
Molecular equations provide a complete overview of a chemical reaction. They include all reactants and products, depicted together as intact molecules. These equations help visualize what happens when these molecules interact.
Let's take a simple example: When manganese (Mn) reacts with dilute sulfuric acid (H\(_2\)SO\(_4\)), the equation looks like this:
  • Reactants: Mn and H\(_2\)SO\(_4\)
  • Products: Manganese sulfate (MnSO\(_4\)) and hydrogen gas (H\(_2\))
The balanced equation is:\[ \text{Mn} + \text{H}_2\text{SO}_4 \rightarrow \text{MnSO}_4 + \text{H}_2 \] The molecular equation shows you every element involved and ensures the number of each type of atom is the same on both sides. Balancing molecular equations is crucial since it respects the law of conservation of mass.
Net Ionic Equations
The net ionic equation focuses only on the ions and molecules directly involved in the chemical change鈥攖hose atoms or molecules that actually participate in the reaction.
Why do we need net ionic equations? They help simplify understanding of reactions by removing all other entities that don't change, i.e., spectator ions.For example, consider the reaction of tin (Sn) with hydrochloric acid (HCl), which results in tin chloride (SnCl\(_2\)) and hydrogen gas (H\(_2\)).
The net ionic equation for this reaction focuses on the tin ions gaining a charge and the release of hydrogen gas:\[ \text{Sn} + 2\text{H}^+ \rightarrow \text{Sn}^{2+} + \text{H}_2 \] This equation highlights the essential transformation from neutral tin atoms to positively charged ions and hydrogen molecules splitting to form hydrogen gas. Net-ionic equations are more focused and are perfect for grasping the substance of a reaction.
Spectator Ions
Spectator ions are the ions that remain unchanged on both sides of a complete ionic equation. They do not participate in the actual chemical change, hence the term 'spectator.' These ions can often be ignored when considering the essential parts of a reaction.

No spectator ions were noted in the presented reactions, meaning each ion in the reactants and products all partake directly in the reaction. For instance, in the reaction of chromium (Cr) with hydrobromic acid (HBr):
  • While bromide (Br\(^-\)) is present in the complete ionic equation, it does not appear in the net ionic version.
  • This is because bromide ions stay the same before and after the reaction.
Identifying spectator ions simplifies the reaction and aids in balancing chemical equations more effectively. It provides a clear, concise view of what is actually changing at the molecular level.

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Most popular questions from this chapter

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 \(\mathrm{Na}^{+}\) are 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?

(a) What volume of \(0.115 \mathrm{M} \mathrm{HClO}_{4}\) solution is needed to neutralize \(50.00 \mathrm{~mL}\) of \(0.0875 \mathrm{M} \mathrm{NaOH} ?\) (b) What volume of \(0.128 \mathrm{M} \mathrm{HCl}\) is needed to neutralize \(2.87 \mathrm{~g}\) of \(\mathrm{Mg}(\mathrm{OH})_{2} ?(\mathbf{c})\) If \(25.8 \mathrm{~mL}\) of an \(\mathrm{AgNO}_{3}\) solution is needed to precipitate all the Cl \(^{-}\) ions in a \(785-m g\) sample of \(K C l\) (forming \(\left.A g C l\right),\) what is the molarity of the \(\mathrm{AgNO}_{3}\) solution? (d) If \(45.3 \mathrm{~mL}\) of a 0.108 \(M\) HCl solution is needed to neutralize a solution of \(\mathrm{KOH}\), how many grams of KOH must be present in the solution?

A \(1.248-g\) sample of limestone rock is pulverized and then treated with \(30.00 \mathrm{~mL}\) of \(1.035 \mathrm{M}\) HCl solution. The excess acid then requires \(11.56 \mathrm{~mL}\) of \(1.010 \mathrm{M} \mathrm{NaOH}\) for neutralization. Calculate the percentage by mass of calcium carbonate in the rock, assuming that it is the only substance reacting with the HCl solution.

Because the oxide ion is basic, metal oxides react readily with acids. (a) Write the net ionic equation for the following reaction: $$ \mathrm{FeO}(s)+2 \mathrm{HClO}_{4}(a q) \longrightarrow \mathrm{Fe}\left(\mathrm{ClO}_{4}\right)_{2}(a q)+\mathrm{H}_{2} \mathrm{O}(l) $$ (b) Based on the equation in part (a), write the net ionic equation for the reaction that occurs between \(\mathrm{NiO}(s)\) and an aqueous solution of nitric acid.

Some sulfuric acid is spilled on a lab bench. You can neutralize the acid by sprinkling sodium bicarbonate on it and then mopping up the resulting solution. The sodium bicarbonate reacts with sulfuric acid according to: $$ \begin{aligned} 2 \mathrm{NaHCO}_{3}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \longrightarrow \mathrm{Na}_{2} \mathrm{SO}_{4}(a q)+& \\ 2 \mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{CO}_{2}(g) \end{aligned} $$ Sodium bicarbonate is added until the fizzing due to the formation of \(\mathrm{CO}_{2}(g)\) stops. If \(27 \mathrm{~mL}\) of \(6.0 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\) was spilled, what is the minimum mass of \(\mathrm{NaHCO}_{3}\), that must be added to the spill to neutralize the acid?

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