6.4: Classifying Chemical Reactions (Acids and Bases) (Problems)
- Page ID
- 106786
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)PROBLEM \(\PageIndex{1}\)
Complete and balance the following acid-base equations:
- HCl gas reacts with solid Ca(OH)2(s).
- A solution of Sr(OH)2 is added to a solution of HNO3.
- Answer a
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\(\ce{2HCl}(g)+\ce{Ca(OH)2}(s)\rightarrow \ce{CaCl2}(s)+\ce{2H2O}(l)\)
- Answer b
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\(\ce{Sr(OH)2}(aq)+\ce{2HNO3}(aq)\rightarrow \ce{Sr(NO3)2}(aq)+\ce{2H2O}(l)\)
PROBLEM \(\PageIndex{2}\)
Complete and balance the following acid-base equations:
- A solution of HClO4 is added to a solution of LiOH.
- Aqueous H2SO4 reacts with NaOH.
- Ba(OH)2 reacts with HF gas.
- Answer a
-
\(\ce{HClO4}(aq)+\ce{LiOH}(aq)\rightarrow \ce{LiClO4}(aq)+\ce{H2O}(l)\)
- Answer b
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\(\ce{H2SO4}(aq)+\ce{2NaOH}(aq)\rightarrow \ce{Na2SO4}(aq)+\ce{2H2O}(l)\)
- Answer c
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\(\ce{2HF}(aq)+\ce{Ba(OH)2}(aq)\rightarrow \ce{BaF2}(aq)+\ce{2H2O}(l)\)
- Click here to see a video of the solution
-
(This video is improperly titled, but is the correct problem solution)
PROBLEM \(\PageIndex{3}\)
Complete and balance the equations for the following acid-base neutralization reactions. If water is used as a solvent, write the reactants and products as aqueous ions. In some cases, there may be more than one correct answer, depending on the amounts of reactants used.
- \(\ce{Mg(OH)2}(s)+\ce{HClO4}(aq)\rightarrow \)
- \(\ce{SO3}(g)+\ce{H2O}(l)\rightarrow \) (assume an excess of water and that the product dissolves)
- \(\ce{SrO}(s)+\ce{H2SO4}(l)\rightarrow \)
- Answer a
-
\(\ce{Mg(OH)2}(s)+\ce{2HClO4}(aq)\rightarrow \ce{Mg^2+}(aq)+\ce{2ClO4-}(aq)+\ce{2H2O}(l)\)
- Answer b
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\(\ce{SO3}(g)+\ce{2H2O}(l)\rightarrow \ce{H3O+}(aq)+\ce{HSO4-}(aq)\), (a solution of H2SO4)
- Answer c
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\(\ce{SrO}(s)+\ce{H2SO4}(l)\rightarrow \ce{SrSO4}(s)+\ce{H2O}\)
PROBLEM \(\PageIndex{4}\)
Complete and balance the equations of the following reactions, each of which could be used to remove hydrogen sulfide from natural gas:
- \(\ce{Ca(OH)2}(s)+\ce{H2S}(g) \rightarrow\)
- \(\ce{Na2CO3}(aq)+\ce{H2S}(g)\rightarrow \)
- Answer
-
\(\ce{Ca(OH)2}(s)+\ce{H2S}(g)\rightarrow \ce{CaS}(s)+\ce{2H2O}(l)\)
- Answer
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\(\ce{Na2CO3}(aq)+\ce{H2S}(g)\rightarrow \ce{Na2S}(aq)+\ce{CO2}(g)+\ce{H2O}(l)\)
PROBLEM \(\PageIndex{5}\)
Calcium cyclamate Ca(C6H11NHSO3)2 is an artificial sweetener used in many countries around the world but is banned in the United States. It can be purified industrially by converting it to the barium salt through reaction of the acid C6H11NHSO3H with barium carbonate, treatment with sulfuric acid (barium sulfate is very insoluble), and then neutralization with calcium hydroxide. Write the balanced equations for these reactions.
- Answer
-
\(\ce{2C6H11NHSO3H}+\ce{BaCO3}\rightarrow \ce{Ba(C6H11NHSO3)2}+\ce{H2CO3}\)
\(\ce{Ba(C6H11NHSO3)2}+\ce{H2SO4}\rightarrow \ce{BaSO4}+\ce{2C6H11NHSO3H}\)
\(\ce{2C6H11NHSO3H}+\ce{Ca(OH)2}\rightarrow \ce{Ca(C6H11NHSO3)2}+\ce{2H2O}\)
Contributors
Paul Flowers (University of North Carolina - Pembroke), Klaus Theopold (University of Delaware) and Richard Langley (Stephen F. Austin State University) with contributing authors. Textbook content produced by OpenStax College is licensed under a Creative Commons Attribution License 4.0 license. Download for free at http://cnx.org/contents/85abf193-2bd...a7ac8df6@9.110).
- Adelaide Clark, Oregon Institute of Technology