Notes
Writing and balancing
- Reactants $\rightarrow$ products, with (s), (l), (g), (aq). Diatomic elements: $\text{H}_2$, $\text{N}_2$, $\text{O}_2$, $\text{F}_2$, $\text{Cl}_2$, $\text{Br}_2$, $\text{I}_2$.
- Balance with coefficients only, never subscripts. Mass is conserved.
Reaction types
| Type | Pattern | Example |
|---|---|---|
| synthesis | $\text{A} + \text{B} \rightarrow \text{AB}$ | $2\text{Mg} + \text{O}_2 \rightarrow 2\text{MgO}$ |
| decomposition | $\text{AB} \rightarrow \text{A} + \text{B}$ | $\text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2$ |
| single replacement | $\text{A} + \text{BC} \rightarrow \text{AC} + \text{B}$ | $\text{Zn} + \text{CuSO}_4 \rightarrow \text{ZnSO}_4 + \text{Cu}$ |
| double replacement | $\text{AB} + \text{CD} \rightarrow \text{AD} + \text{CB}$ | $\text{AgNO}_3 + \text{NaCl} \rightarrow \text{AgCl} + \text{NaNO}_3$ |
| combustion | fuel $+ \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O}$ | $\text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O}$ |
Predicting products
- Activity series: a metal replaces only metals below it; metals above H react with acids to make $\text{H}_2$.
- Solubility: $\text{Na}^+$, $\text{K}^+$, $\text{NH}_4^+$ and $\text{NO}_3^-$ compounds are always soluble. Double replacement happens when a precipitate, gas or water forms.
- Net ionic equations leave out spectator ions.
Topics in this unit
Each topic has its own short notes and practice.
Practice
$2\text{Na} + 2\text{H}_2\text{O} \rightarrow 2\text{NaOH} + \text{H}_2$: 2 Na, 4 H and 2 O on each side.
Two compounds swap partners.
Silver is below hydrogen in the activity series.
One compound breaks down into two simpler substances.
AgBr is insoluble. The other pairs swap to form only soluble compounds.
Products contain $2(2) + 2(1) = 6$ O atoms, so $3\text{O}_2$.
Free response
Write your answer first, then open the scoring guide.
Free-response question
In a lab, a student mixes clear, colorless solutions of lead(II) nitrate and potassium iodide. A bright yellow solid forms at once.
(a) Write the balanced molecular equation with state symbols.
Show the answer and scoring guide
$\text{Pb(NO}_3)_2\text{(aq)} + 2\text{KI(aq)} \rightarrow \text{PbI}_2\text{(s)} + 2\text{KNO}_3\text{(aq)}$
Check off the points you earned:
(b) Name the reaction type and identify the yellow solid. Explain how you know which product is the solid.
Show the answer and scoring guide
It is a double replacement reaction. The yellow solid is lead(II) iodide, $\text{PbI}_2$. Potassium and nitrate compounds are always soluble, so $\text{KNO}_3$ stays dissolved; $\text{PbI}_2$ is insoluble, so it precipitates.
Check off the points you earned:
(c) Write the net ionic equation and name the spectator ions.
Show the answer and scoring guide
$\text{Pb}^{2+}\text{(aq)} + 2\text{I}^-\text{(aq)} \rightarrow \text{PbI}_2\text{(s)}$. Spectator ions: $\text{K}^+$ and $\text{NO}_3^-$.
Check off the points you earned: