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Complete and balance the following molecular equations, and then write the net ionic equation for each: (a) \(\mathrm{HBr}(a q)+\mathrm{Ca}(\mathrm{OH})_{2}(a q) \longrightarrow\) (b) \(\mathrm{Cu}(\mathrm{OH})_{2}(s)+\mathrm{HClO}_{4}(a q) \longrightarrow\) (c) \(\mathrm{Al}(\mathrm{OH})_{3}(s)+\mathrm{HNO}_{3}(a q) \longrightarrow\)

Short Answer

Expert verified
(a) Balanced molecular equation: HBr + Ca(OH)鈧 鈫 2 H鈧侽 + CaBr鈧 \\ Net ionic equation: H鈦(aq) + OH鈦(aq) 鈫 H鈧侽(l) (b) Balanced molecular equation: Cu(OH)鈧(s) + 2 HClO鈧(aq) 鈫 2 H鈧侽(l) + Cu(ClO鈧)鈧(aq) \\ Net ionic equation: Cu(OH)鈧(s) + 2 H鈦(aq) 鈫 Cu虏鈦(aq) + 2 H鈧侽(l) (c) Balanced molecular equation: Al(OH)鈧(s) + 3 HNO鈧(aq) 鈫 3 H鈧侽(l) + Al(NO鈧)鈧(aq) \\ Net ionic equation: Al(OH)鈧(s) + 3 H鈦(aq) 鈫 Al鲁鈦(aq) + 3 H鈧侽(l)

Step by step solution

01

(a) Completing and balancing the molecular equation for HBr and Ca(OH)鈧

First, notice that HBr (hydrobromic acid) and Ca(OH)鈧 (calcium hydroxide) are both aqueous solutions. The products formed in this reaction will be water (H鈧侽) and a soluble salt. Since HBr is an acid and Ca(OH)鈧 is a base, this reaction is an acid-base neutralization reaction. The products formed are water and calcium bromide (CaBr鈧): HBr + Ca(OH)鈧 鈫 H鈧侽 + CaBr鈧 To balance the equation, we see that there are 2 bromine atoms and 2 hydroxide ions on the reactants' side. Since there are 2 bromine atoms in CaBr鈧, we only need to multiply H鈧侽 by 2 to balance the hydrogen in OH and HBr, So the balanced equation is: HBr + Ca(OH)鈧 鈫 2 H鈧侽 + CaBr鈧
02

(a) Writing the total and net ionic equation for HBr and Ca(OH)鈧

Now we need to write the total ionic equation by breaking up the soluble substances into their individual ions in the solution: H鈦(aq) + Br鈦(aq) + Ca虏鈦(aq) + 2 OH鈦(aq) 鈫 2 H鈧侽(l) + Ca虏鈦(aq) + 2 Br鈦(aq) By examining the equation, we can identify the spectator ions that remain unchanged, which are Ca虏鈦 and Br鈦. We can then remove them from the equation, yielding the net ionic equation: H鈦(aq) + OH鈦(aq) 鈫 H鈧侽(l)
03

(b) Completing and balancing the molecular equation for Cu(OH)鈧 and HClO鈧

We can see that Cu(OH)鈧 is a solid and the reaction with HClO鈧 (perchloric acid) is another acid-base reaction. The products of this reaction are water and copper(II) perchlorate salt (Cu(ClO鈧)鈧): Cu(OH)鈧(s) + HClO鈧(aq) 鈫 H鈧侽(l) + Cu(ClO鈧)鈧(aq) To balance the equation, we need 2 HClO鈧 molecules to balance the 2 OH鈦 ions in Cu(OH)鈧: Cu(OH)鈧(s) + 2 HClO鈧(aq) 鈫 2 H鈧侽(l) + Cu(ClO鈧)鈧(aq)
04

(b) Writing the total and net ionic equation for Cu(OH)鈧 and HClO鈧

Now we write the total ionic equation: Cu(OH)鈧(s) + 2 H鈦(aq) + 2 ClO鈧勨伝(aq) 鈫 2 H鈧侽(l) + Cu虏鈦(aq) + 2 ClO鈧勨伝(aq) The spectator ions are 2 ClO鈧勨伝 ions. The net ionic equation will be: Cu(OH)鈧(s) + 2 H鈦(aq) 鈫 Cu虏鈦(aq) + 2 H鈧侽(l)
05

(c) Completing and balancing the molecular equation for Al(OH)鈧 and HNO鈧

In this reaction, Al(OH)鈧 is a solid and reacts with HNO鈧 (nitric acid) in another acid-base reaction. This reaction produces water and aluminum nitrate salt (Al(NO鈧)鈧): Al(OH)鈧(s) + HNO鈧(aq) 鈫 H鈧侽(l) + Al(NO鈧)鈧(aq) We have 3 OH鈦 in Al(OH)鈧(s) and that requires 3 HNO鈧 molecules: Al(OH)鈧(s) + 3 HNO鈧(aq) 鈫 3 H鈧侽(l) + Al(NO鈧)鈧(aq)
06

(c) Writing the total and net ionic equation for Al(OH)鈧 and HNO鈧

The total ionic equation for this reaction is: Al(OH)鈧(s) + 3 H鈦(aq) + 3 NO鈧冣伝(aq) 鈫 3 H鈧侽(l) + Al鲁鈦(aq) + 3 NO鈧冣伝(aq) The spectator ions are 3 NO鈧冣伝 ions. The net ionic equation is: Al(OH)鈧(s) + 3 H鈦(aq) 鈫 Al鲁鈦(aq) + 3 H鈧侽(l)

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Key Concepts

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

Net ionic equations
Net ionic equations are simplified versions of chemical equations that show only the chemical species involved in the reaction. This type of equation removes spectator ions, which do not change during the reaction process. By stripping away these unreacted components, the net ionic equation more clearly illustrates the essence of the chemical change that occurs.

Determining the net ionic equation involves:
  • Writing the balanced molecular equation.
  • Breaking down all aqueous compounds into their respective ions to form the complete ionic equation.
  • Removing the spectator ions (ions that appear unchanged on both sides of the equation).
For instance, in the reaction between hydrobromic acid (HBr) and calcium hydroxide (Ca(OH)鈧), the net ionic equation is \[\text{H}^+(aq) + \text{OH}^- (aq) \rightarrow \text{H}_2\text{O}(l)\]. This equation focuses on the key process of forming water from hydrogen and hydroxide ions.
Balancing chemical equations
Balancing chemical equations is a crucial step in representing chemical reactions accurately. It ensures that the law of conservation of mass is maintained, meaning the number of atoms of each element is the same on both sides of the equation. This task requires:
  • Identifying each compound involved in the reaction.
  • Counting the number of atoms for each element on both sides of the equation.
  • Adding coefficients to the reactants or products to equate the number of atoms, avoiding alteration of the chemical formulas themselves.
For example, the initial equation representing the reaction between HBr and Ca(OH)鈧:\[\text{HBr} + \text{Ca(OH)}_2 \rightarrow \text{H}_2\text{O} + \text{CaBr}_2\]is balanced by ensuring there are equal numbers of H, Br, and OH species through careful adjustment of coefficients yielding:\[2 \text{HBr} + \text{Ca(OH)}_2 \rightarrow 2 \text{H}_2\text{O} + \text{CaBr}_2\].

This balance reveals the stoichiometric relationships between the reactants and products, which is essential for quantitative chemistry.
Spectator ions
In many chemical reactions, especially those occurring in aqueous solutions, certain ions do not participate actively in the chemical change. These are called spectator ions. They simply "watch" the reaction without undergoing any change in oxidation state or composition.

Spectator ions appear in the complete ionic equation on both sides, unchanged. Identifying them helps simplify chemical equations. For instance, in the reaction of HBr and Ca(OH)鈧, the complete ionic equation contains calcium ions (\(\text{Ca}^{2+}\)) and bromide ions (\(\text{Br}^-\)) as spectator ions, because they do not take part in forming water.

Once these inactive participants are removed, what remains is the net ionic equation. This simplification is useful not only for understanding chemical processes better, but also aids in calculations, such as determining concentrations of reacting species.

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Most popular questions from this chapter

We have learned in this chapter that many ionic solids dissolve in water as strong electrolytes; that is, as separated ions in solution. Which statement is most correct about this process? (a) Water is a strong acid and therefore is good at dissolving ionic solids. (b) Water is good at solvating ions because the hydrogen and oxygen atoms in water molecules bear partial charges. (c) The hydrogen and oxygen bonds of water are easily broken by ionic solids.

Classify each of the following substances as a nonelectrolyte, weak electrolyte, or strong electrolyte in water: (a) HF, (b) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}\) (benzoicacid), (c) \(\mathrm{C}_{6} \mathrm{H}_{6}\) (benzene), (d) \(\mathrm{CoCl}_{3}\) (e) \(\mathrm{AgNO}_{3}\)

The commercial production of nitric acid involves the following chemical reactions: $$ \begin{aligned} 4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) & \longrightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g) \\ 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) & \longrightarrow 2 \mathrm{NO}_{2}(g) \\ 3 \mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) & \longrightarrow 2 \mathrm{HNO}_{3}(a q)+\mathrm{NO}(g) \end{aligned} $$ (a) Which of these reactions are redox reactions? (b) In each redox reaction identify the element undergoing oxidation and the element undergoing reduction. (c) How many grams of ammonia must you start with to make \(1000.0 \mathrm{~L}\) of a \(0.150 \mathrm{M}\) aqueous solution of nitric acid? Assume all the reactions give \(100 \%\) yield.

The metal cadmium tends to form Cd \(^{2+}\) ions. The following observations are made: (i) When a strip of zinc metal is placed in \(\mathrm{CdCl}_{2}(a q),\) cadmium metal is deposited on the strip. (ii) When a strip of cadmium metal is placed in \(\mathrm{Ni}\left(\mathrm{NO}_{3}\right)_{2}(a q),\) nickel metal is deposited on the strip. (a) Write net ionic equations to explain each of the preceding observations. (b) Which elements more closely define the position of cadmium in the activity series? (c) What experiments would you need to perform to locate more precisely the position of cadmium in the activity series?

Using the activity series (Table 4.5 ), write balanced chemical equations for the following reactions. If no reaction occurs, write NR. (a) Nickel metal is added to a solution of copper(II) nitrate, (b) a solution of zinc nitrate is added to a solution of magnesium sulfate, (c) hydrochloric acid is added to gold metal, (d) chromium metal is immersed in an aqueous solution of cobalt(II) chloride, (e) hydrogen gas is bubbled

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