Chapter 5: Problem 58
Write balanced molecular, ionic, and net ionic equations for the reactions of the following metals with hydrochloric acid to give hydrogen plus the metal ion in solution. (a) Cobalt (gives \(\mathrm{Co}^{2+}\) ) (b) Cesium (gives \(\left.\mathrm{Cs}^{+}\right)\) (c) Zinc (gives \(\mathrm{Zn}^{2+}\) )
Short Answer
Expert verified
The net ionic equations for the reactions with HCl are: Cobalt, \(\mathrm{Co(s) + 2H^+(aq) \rightarrow Co^{2+}(aq) + H_2(g)}\); Cesium, \(\mathrm{2Cs(s) + 2H^+(aq) \rightarrow 2Cs^+(aq) + H_2(g)}\); Zinc, \(\mathrm{Zn(s) + 2H^+(aq) \rightarrow Zn^{2+}(aq) + H_2(g)}\).
Step by step solution
01
Write the balanced molecular equation for Cobalt
The reaction of cobalt with hydrochloric acid can be represented by the molecular formula: \[ \mathrm{Co (s) + 2HCl (aq) \rightarrow CoCl_2 (aq) + H_2 (g)} \] The cobalt metal (Co) reacts with hydrochloric acid (HCl) to form cobalt chloride (CoCl_2) and hydrogen gas (H_2). To balance the equation, we need two moles of HCl for every mole of Co, leading to one mole each of CoCl_2 and H_2.
02
Write the ionic equation for Cobalt
In the ionic equation, we write all soluble substances as their ions: \[ \mathrm{Co (s) + 2H^+ (aq) + 2Cl^- (aq) \rightarrow Co^{2+} (aq) + 2Cl^- (aq) + H_2 (g)} \] Solid cobalt remains as a solid, while HCl, being a strong acid, dissociates completely into H\textsuperscript{+} and Cl\textsuperscript{-} ions.
03
Determine the net ionic equation for Cobalt
The net ionic equation only includes the components that undergo a change in state or composition. The spectator ion, \( \mathrm{Cl^-} \), which does not change, is omitted: \[ \mathrm{Co (s) + 2H^+ (aq) \rightarrow Co^{2+} (aq) + H_2 (g)} \]
04
Write the balanced molecular equation for Cesium
Cesium's reaction with hydrochloric acid is represented as: \[ \mathrm{Cs (s) + HCl (aq) \rightarrow CsCl (aq) + \frac{1}{2} H_2 (g)} \] However, we cannot have a fraction of a molecule in the balanced equation, so we multiply everything by 2 to get whole numbers: \[ \mathrm{2 Cs (s) + 2 HCl (aq) \rightarrow 2 CsCl (aq) + H_2 (g)} \]
05
Write the ionic equation for Cesium
The ionic equation for cesium and hydrochloric acid is written by showing all soluble substances as their respective ions: \[ \mathrm{2 Cs (s) + 2 H^+ (aq) + 2 Cl^- (aq) \rightarrow 2 Cs^+ (aq) + 2 Cl^- (aq) + H_2 (g)} \]
06
Determine the net ionic equation for Cesium
We remove the spectator ions to get the net ionic equation: \[ \mathrm{2 Cs (s) + 2 H^+ (aq) \rightarrow 2 Cs^+ (aq) + H_2 (g)} \]
07
Write the balanced molecular equation for Zinc
Zinc reacts with hydrochloric acid according to the following molecular equation: \[ \mathrm{Zn (s) + 2 HCl (aq) \rightarrow ZnCl_2 (aq) + H_2 (g)} \] This equation is balanced with one mole of zinc reacting with two moles of hydrochloric acid to produce one mole of zinc chloride and one mole of hydrogen gas.
08
Write the ionic equation for Zinc
To write the ionic equation, all the components in aqueous solution are shown as ions: \[ \mathrm{Zn (s) + 2 H^+ (aq) + 2 Cl^- (aq) \rightarrow Zn^{2+} (aq) + 2 Cl^- (aq) + H_2 (g)} \]
09
Determine the net ionic equation for Zinc
The net ionic equation, after removing the chloride ions which are spectators, becomes: \[ \mathrm{Zn (s) + 2 H^+ (aq) \rightarrow Zn^{2+} (aq) + H_2 (g)} \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Balanced Chemical Equations
A balanced chemical equation shows the identity of the compounds involved in a reaction, including the substances reacting (reactants) and the substances produced (products). These are written with the exact number of atoms of each element conserved on both sides of the equation. Balancing equations is crucial to accurately represent the law of conservation of mass, stating that matter is neither created nor destroyed during a reaction.
For example, the balanced equation for the reaction of cobalt with hydrochloric acid is: �...�,��,�...�,��,�...�,��,�...�,��,�...�,��,�...�,��,�...�,��,�...�,��,��,��,���...��,���...��,���...��,���...��,���...��,���...��,���...��,���...��,��,��,�...�,�...�,�...�,�...�,�...�,���...��,�...�,�...�,��,�...�,��,�...�,��,�...�,��,�...�,��,�...�,��,���...��,��,��.�� �...�,���...��,��,�...�,���...��,�...�,��,�...�,������...��,��,��,�...�,��,��,���...��,��,��,��,��,��,��.�� �...�,��,��.�� �...�,��,��,��,��,��,��.�� �...�,��,��.���...�,��,��,��,��,��,��.�� �...�,��,��.�� �...�,��,��,��,��,��,��.�� �...�,��,��.�� �...�,��,��,��,��,��,��.�� �...�,��,��.�� �...�,��,��,��,��,��,��.�� �...�,��,��.�� �...�,��,��,��,��,��,��.�� �...�,��,��.�� �...�,��,��,��,��,��,��.�� �...�,�...�,�...�,���...��,���...��,��,��,����.�� �...�,��,��.�� �...�,���...��,��,���...��,��,����.�� �...�,���...��,��,���...��,��,����.�� �...�,��,��,��,��,��,��.�� �...�,��,��.�� �...�,��,����.�� �...�,�...�,�...�,���...��,���...��,��,��,��.�� �...�,��,��.�� �...�,��,��,��,��,��,��.�� �...�,��,��.�� �...�,��,��,��,��,��,��.�� �...�,��,��.�� �...�,��,��,��,��,��,��.�� �...�,��,��.���...�,��,��,��,��,��,���...��,��,��.�� �...�,��,��.�� �...�,��,��,����.�� �...�,�...�,�...�,���...��,���...��,����.�� �...�,��,��.�� �...�,��,��,��,��,��,��.�� �...�,��,��.�� �...�,��,��,��,��,��,��.�� �...�,��,��.�� �...�,��,��,��,��,��,��.�� �...�,��,��.���...�,��,��,��,��,��,���...��,��,��.�� �...�,��,��.�� �...�,��,��,����.�� �...�,��,��,��,��,��,��.�� �...�,��,��.�� �...�,��,��,��,��,��,��.�� �...�,��,��.�� �...�,��,��,��,��,��,��.�� �...�,��,��.���...�,��,��,��,��,��,��.�� �...�,��,��.�� �...�,��,����.�� �...�,�...�,�...�,���...��,���...��,��,��,��.�� �...�,��,��.�� �...�,��,��,��,��,��,��.�� Properly balancing chemical equations is the foundation for understanding the stoichiometry of a reaction, which deals with the quantitative relationships between reactants and products in a chemical reaction. This is essential for predicting the quantities of substances needed or produced in a chemical process.
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Molecular Equations
Molecular equations provide a straightforward representation of a chemical reaction using the formulas of the compounds in their standard states. This is the type of equation most commonly written in chemistry and seen in textbooks.
The molecular equation for cesium reacting with hydrochloric acid looks like this: �...�,��,�...�,��,�...�,��,�...�,��,�...�,��,�...�,��,�...�,��,���...��,��,��,��,��,��,��.�� �...�,��,��.�� �...�,��,��,��,��,��,��.�� Unlike the net ionic equation, the molecular equation does not explicitly illustrate the ionic nature of the aqueous compounds. It assumes they remain intact as molecules throughout the reaction. However, this simplification often makes it easier to quickly comprehend the overall chemical change.
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Ionic Equations
Ionic equations take the molecular equation a step further by representing soluble ionic compounds and strong acids as dissociated ions. This depiction is more accurate for aqueous solutions, where such dissociation occurs. Ionic equations help us see the actual forms of substances when a chemical reaction takes place in water.
Here's an example with zinc reacting with hydrochloric acid: �...�,��,�...�,��,�...�,��,�...�,��,�...�,��,�...�,��,�...�,��,���...��,��,��,��,��,��,��.�� �...�,��,��.�� �...�,��,��,��,��,��,��.�� While writing ionic equations, we present the state of each substance: (s) for solid, (l) for liquid, (g) for gas, and (aq) for aqueous, meaning dissolved in water. This clarifies the phases involved and sets the stage for identifying spectator ions and deriving the net ionic equation.
Here's an example with zinc reacting with hydrochloric acid: �...�,��,�...�,��,�...�,��,�...�,��,�...�,��,�...�,��,�...�,��,���...��,��,��,��,��,��,��.�� �...�,��,��.�� �...�,��,��,��,��,��,��.�� While writing ionic equations, we present the state of each substance: (s) for solid, (l) for liquid, (g) for gas, and (aq) for aqueous, meaning dissolved in water. This clarifies the phases involved and sets the stage for identifying spectator ions and deriving the net ionic equation.
Reaction with Hydrochloric Acid
Understanding how metals react with hydrochloric acid is quite significant as this is a typical single-displacement reaction, where a metal displaces hydrogen from the acid. This specific type of acid-base reaction produces hydrogen gas and salt of the respective metal.
The general form of such a reaction follows the pattern: �...�,��,�...�,��,�...�,��,�...�,��,�...�,��,�...�,��,�...�,��,���...��,��,��,��,��,��,��.�� �...�,��,��.�� �...�,��,��,��,��,��,��.�� Hydrochloric acid ( �...�,��,�...�,��,�...�,��,�...�,��,�...�,��,�...�,��,�...�,��,���...��,��,��,��,��,��,��.�� The metal involved determines the specific products formed and their solubility, which then influences whether the full equation should be written as ionic or molecular.
The general form of such a reaction follows the pattern: �...�,��,�...�,��,�...�,��,�...�,��,�...�,��,�...�,��,�...�,��,���...��,��,��,��,��,��,��.�� �...�,��,��.�� �...�,��,��,��,��,��,��.�� Hydrochloric acid ( �...�,��,�...�,��,�...�,��,�...�,��,�...�,��,�...�,��,�...�,��,���...��,��,��,��,��,��,��.�� The metal involved determines the specific products formed and their solubility, which then influences whether the full equation should be written as ionic or molecular.
Hydrogen Gas Production
When a metal reacts with an acid, like hydrochloric acid, one of the signature products is hydrogen gas ( �...�,��,�...�,��,�...�,��,�...�,��,�...�,��,�...�,��,�...�,��,���...��,��,��,��,��,��,��.�� �...�,��,��.�� �...�,��,��,��,��,��,��.�� The evolution of hydrogen gas is an indication that a chemical reaction has taken place. For instance, the reaction of cobalt with hydrochloric acid produces hydrogen gas as one of the products. This gas can be tested with a lit splint, which will produce a 'pop' sound if hydrogen is present. Observing the production of hydrogen gas not only confirms the activity of the metal but also helps in verifying that the reaction is proceeding correctly. Moreover, accurately predicting the generation of hydrogen is essential for maintaining safety in the lab environment as hydrogen is highly flammable.