6: Chemical Reactions and Equations
- Page ID
- 506164
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)The purpose of this experiment is to:
- Run several chemical reactions and record observations.
- Balance chemical equations.
- Predict the products of single- and double-replacement reactions.
INTRODUCTION
Chemical changes produce new materials. In a chemical change, atoms are rearranged as reactants (the starting materials) and are converted into products (the new materials). Chemical equations can represent chemical changes. Consider the conversion of hydrogen and oxygen into water:
H2 + O2 -> H2O
In a chemical reaction, matter cannot be created or destroyed, meaning the same number and type of atoms must be shown on each side of the equation. To illustrate this, we must balance the equation by adding a whole number coefficient in front of each element or compound. We can balance the above equation as follows:
2 H2 + O2 —-> 2 H2O
The total number of atoms on each side of the equation can be found by multiplying the added coefficient by the given subscript. On the left side of the equation, we see four hydrogens (2x2) and two oxygens (1x2). The right side of the equation also has four hydrogens (2 × 2) and two oxygens (2 × 1). Since this equation has the same number of atoms on each side, it is balanced. Refer to the course lectures and notes for examples and detailed steps on balancing equations.
In this experiment, you will explore three reaction types: decomposition, single, and double displacement. A decomposition reaction occurs when one reactant forms two or more products. An example would be the conversion of sodium azide into sodium metal and nitrogen gas, as shown below:
2 NaN3 (s) —-> 2 Na (s) + 3 N2 (g)
In a single displacement reaction, one element replaces another in a compound. These reactions often run in aqueous solutions. Consider the reaction of magnesium metal with aluminum chloride:
3 Mg (s) + 2 AlCl3 (aq) —-> 3 MgCl2 (aq) + 2 Al (s)
In this reaction, magnesium replaces aluminum with aluminum chloride.
Like single displacement reactions, double displacement reactions also occur in aqueous solutions, and cations and anions exchange places according to the following generic equation:
AX + BY —-> AY + BX
In many cases, when two aqueous solutions are mixed, a solid, known as a precipitate, forms, as shown by the reaction between sodium bromide and silver nitrate:
NaBr (aq) + AgNO3 (aq) —-> NaNO3 (aq) + AgBr (s)
To predict the products of a double displacement reaction, write the four ions present in the solution (in this case, Na+, Br-, Ag+, and NO3-) and pair opposite ions together, writing the formula for the ionic compound. One quick way to identify a precipitate (if it forms) is by looking at the two cations. The precipitate typically forms from the heavier metal cation. In this reaction, silver is heavier than sodium. A more formal method for determining whether a precipitate will form when two aqueous solutions are mixed is by examining a list of solubility guidelines provided in the course textbook.