Notes
Physical vs chemical change
A chemical change breaks and forms bonds and makes new substances (gas produced, color change, precipitate, temperature change). A physical change, like dissolving sugar or melting ice, changes only intermolecular attractions. Dissolving ionic solids is a gray area because ionic bonds break.
Equations
- Balanced equations conserve atoms and charge.
- Net ionic equations show only species that change. Strong electrolytes (soluble ionic compounds, strong acids) are written as ions; spectator ions are removed. Example: $Ag^+(aq) + Cl^-(aq) \rightarrow AgCl(s)$.
- All sodium, potassium, ammonium and nitrate salts are soluble.
Stoichiometry
Grams → moles → mole ratio from the balanced equation → moles → grams.
- Limiting reactant: the one that runs out first and sets the maximum product (theoretical yield). Percent yield $\displaystyle = \frac{\text{actual}}{\text{theoretical}} \times 100$.
- Titration: at the equivalence point, moles of titrant match the moles of analyte by the reaction ratio.
Types of reactions
- Acid-base (Brønsted-Lowry): proton ($\mathrm{H^+}$) transfer. The acid donates; the base accepts.
- Redox: electron transfer. Oxidation is loss of electrons (oxidation number increases); reduction is gain.
- Precipitation: two solutions form an insoluble solid.
- Oxidation number rules: elements are 0; O is usually −2; H is usually +1; the sum equals the overall charge.
Topics in this unit
Each topic has its own short notes and practice.
Practice
Rusting forms a new substance (iron oxide).
$\mathrm{PbI_2}$ precipitates; $\mathrm{K^+}$ and $\mathrm{NO_3^-}$ are spectators.
The ratio of $\mathrm{H_2O}$ to $\mathrm{O_2}$ is 2 : 1.
$4.0/2 = 2.0$ for $\mathrm{H_2}$ vs $3.0/1 = 3.0$ for $\mathrm{O_2}$; $\mathrm{H_2}$ is smaller.
$x + 4(-2) = -1$, so $x = +7$.
Zn loses two electrons.
Moles HCl $= 0.00200$; $\displaystyle \frac{0.00200}{0.0250} = 0.0800$ M.
Water donates a proton to ammonia.
Free response
Write your answer first, then open the scoring guide.
Free-response question
A student reacts 5.00 g of magnesium with excess hydrochloric acid: $Mg(s) + 2HCl(aq) \rightarrow MgCl_2(aq) + H_2(g)$.
(a) Calculate the theoretical yield of $\mathrm{H_2}$ in grams.
Show the answer and scoring guide
$\displaystyle \frac{5.00}{24.31} = 0.206$ mol Mg, which gives 0.206 mol $\mathrm{H_2}$. $0.206 \times 2.016 = 0.415$ g $\mathrm{H_2}$.
Check off the points you earned:
(b) The student collected 0.380 g of $\mathrm{H_2}$. Calculate the percent yield.
Show the answer and scoring guide
$\displaystyle \frac{0.380}{0.415} \times 100 \approx 91.6\%$.
Check off the points you earned:
(c) Identify the species oxidized and reduced.
Show the answer and scoring guide
Mg is oxidized (0 to +2). $\mathrm{H^+}$ is reduced (+1 to 0, forming $\mathrm{H_2}$).
Check off the points you earned: