Unit 4: Chemical Reactions

Physical versus chemical change

A chemical change produces new substances with new chemical identities. Observable clues include a lasting color change, gas evolution (from something other than simple boiling), formation of a precipitate, a large temperature change from the reaction itself, light emission, or a new odor. Those clues are suggestive, not proof; the decisive idea is reorganization at the molecular level (bond breaking and forming).
A physical change alters state, size, shape, or mixing without creating a new chemical species: melting, boiling, dissolving sugar in water (the sucrose molecules remain intact), or crushing a sample. Dissolving an ionic compound in water is still often grouped with “physical” solution formation in introductory courses, even though the ions separate from the crystal—a distinction worth keeping straight when you discuss conductivity and equilibrium later.

Example. Bubbles appear both when water boils and when acid is added to a carbonate. Explain why the bubbles alone cannot distinguish physical and chemical change.

Boiling produces water vapor, so molecular identity is retained. Acid-carbonate reaction produces carbon dioxide, a different substance. Gas formation is an observation; identifying the gas or showing the change in composition establishes which process occurred.

Representing and balancing reactions

A chemical equation lists reactants (left) and products (right), usually separated by a single arrow $(\longrightarrow)$ for a reaction treated as one-way in stoichiometry, or $(\rightleftharpoons)$ when equilibrium matters (Unit 7). State symbols clarify what you are counting:

$(s)$ solid, $(l)$ liquid, $(g)$ gas, $(aq)$ dissolved in water (aqueous).

If an element is a component of the reaction ALONE, they are written in their standard/naturally-occuring form (e.g. $\text{O}_2(g)$ , $\text{Na}(s)$ , $\text{Br}_2(l)$ ). Otherwise, they can take on any form. Coefficients are the smallest integers (or a set of integers) consistent with conservation of atoms: they give mole ratios for limiting-reactant work, titrations, and gas-law stoichiometry. Make sure that on each side of a reaction, you start and end with the same number of atoms/moles.

Balancing by inspection (non-redox and simple cases)

Balance elements that appear in only one reactant and one product first (often metals or central atoms).
Treat unchanged polyatomic ions (such as $\text{NO}_3^-$ or $\text{SO}_4^{2-}$ ) as a unit if they appear intact on both sides (Make sure to include them in brackets, otherwise they do not count as polyatomic ions).
Balance hydrogen and oxygen last when they appear in several compounds (common in combustion and acid–base).
If you temporarily need a fractional coefficient to balance an element (e.g. $\frac{1}{2}\text{O}_2$ ), multiply the entire equation by the denominator so all coefficients are integers.

The balanced equation conserves mass (atom counts). For ionic reactions in solution, you may also write:

Molecular equation — Molecular formulas are formulas as written in the bottle/usually stated in the problem (e.g. $\text{AgNO}_3(aq) + \text{NaCl}(aq)$ )
Total ionic equation — You treat strong electrolytes as separated ions, while solids, liquids, weak electrolytes, and gases usually undissociated (Check solubility rules below for electrolytes; strong electrolytes are soluble in water)
Net ionic equation — Write the total ionic equation but cancel spectator ions that do not change (Basically anything that can cancel on both sides) and only species that actually react

Redox reactions in acidic or basic solution often needs the half-reaction method (below) because oxidation numbers change and electron and charge balance are not handled by atom counting alone.

Example. A student balances hydrogen combustion by changing $\mathrm{H_2O}$ to $\mathrm{H_2O_2}$ . Explain the error and give the balanced equation for making water.

Changing a subscript changes the product’s identity to hydrogen peroxide. Coefficients change amounts without changing identities: $2\mathrm{H_2}+\mathrm{O_2}\rightarrow2\mathrm{H_2O}$ . Both sides now contain four H atoms and two O atoms.

Stoichiometry, limiting reactants, and yield

Once an equation is balanced, the coefficients are mole ratios: a conversion factor between any two species in the reaction. Every quantitative reaction problem is some version of the same three-step path:

Convert what you are given into moles (using molar mass for a mass, $M\times V$ for a solution, or the ideal gas law $PV=nRT$ for a gas).
Use the mole ratio from the balanced equation to cross over to the species you want.
Convert moles of the wanted species back into the requested unit (grams, liters of solution, liters of gas, particles).

\text{given quantity} \longrightarrow \text{mol given} \xrightarrow{\text{mole ratio}} \text{mol wanted} \longrightarrow \text{wanted quantity}.

Limiting and excess reactants

When amounts of two or more reactants are given, one usually runs out first and caps how much product forms—the limiting reactant. The others are excess reactants and have leftover material when the reaction stops. To find the limiting reactant, convert each reactant to moles and divide by its coefficient; the smallest of these ratios marks the limiting reactant. Equivalently, pick either reactant, compute how much of the other it would require, and compare to how much is actually present.

At the particle level, the limiting reactant runs out first. Some particles of the excess reactant remain because there are no longer enough particles of the other reactant to react in the required ratio.

Example. For $\mathrm{N_2+3H_2\rightarrow2NH_3}$ , mix $2.00\ \mathrm{mol}$ nitrogen with $3.00\ \mathrm{mol}$ hydrogen. A student calls nitrogen limiting because fewer moles were supplied. Find the error and the remaining reactant amount.

Compare amounts divided by coefficients: $2.00/1=2.00$ for nitrogen and $3.00/3=1.00$ for hydrogen. Hydrogen allows less reaction and is limiting. It consumes $1.00\ \mathrm{mol}$ nitrogen, leaving $1.00\ \mathrm{mol}$ nitrogen and producing $2.00\ \mathrm{mol}$ ammonia. Raw mole counts cannot be compared without the reaction ratio.

Percent yield

The theoretical yield is the maximum product predicted from the limiting reactant by stoichiometry. The actual yield is what is recovered in the lab, which is almost always smaller because of side reactions, incomplete reactions, transfer losses, or impure reactants. The two are compared with

\text{percent yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100\%.

A yield above $100\%$ usually signals a measurement problem, such as a product that is still wet with solvent or contaminated with leftover reactant.

Example. A reaction has a theoretical yield of $1.20\ \mathrm{g}$ , but the collected solid weighs $1.32\ \mathrm{g}$ . Is the reaction more than completely successful? Give a plausible explanation and its effect on the calculated yield.

The apparent yield is $1.32/1.20\times100\%=110\%$ . This does not mean more pure product formed than stoichiometry allows. Retained water or another contaminant could contribute to the measured mass, making the numerator too large. Drying to constant mass can test for retained solvent; an incorrect theoretical-yield calculation is another possibility to check.

Gravimetric and volumetric analysis

Two common quantitative experiments rest entirely on this stoichiometry:

Gravimetric analysis isolates a product (often a filtered, dried precipitate) and weighs it; the mass of the pure solid is converted through the mole ratio back to the amount of the unknown ion in the original sample.
Titration (volumetric analysis) delivers a measured volume of a solution of known concentration (the titrant) until it exactly reacts with the analyte at the equivalence point. From $n=MV$ and the mole ratio you solve for the unknown concentration or amount. Acid–base titrations are explored fully in Unit 8; redox titrations follow the same bookkeeping using the balanced electron-transfer equation.

Example. A $0.500\ \mathrm{g}$ impure chloride sample produces $0.574\ \mathrm{g}$ dry AgCl after adding excess silver nitrate. Find the chloride mass percent using molar masses $143.3$ for AgCl and $35.45\ \mathrm{g/mol}$ for Cl. Explain the effect of incomplete precipitation.

The precipitate represents $0.574/143.3=0.00401\ \mathrm{mol}$ chloride because AgCl contains one chloride per formula unit. Chloride mass is $0.00401(35.45)=0.142\ \mathrm{g}$ , giving $28.4\%$ chloride by mass. Incomplete precipitation leaves some chloride unmeasured and biases the result low. The whole precipitate mass cannot be counted as original chloride because much of it is added silver.

Double-displacement (metathesis) and precipitation

In a double-displacement reaction, cations and anions exchange partners, often in solution:

\text{AB}(aq) + \text{CD}(aq) \longrightarrow \text{AD} (aq) + \text{CB} (s)

If one combination is insoluble (in the example above it is $CB$ ), it forms a precipitate, a solid that may appear as cloudiness, flecks, or a settled solid at the bottom of the vessel. Different precipitates will have different colors, which may be useful in determining the contents of the reaction. If all ionic products remain soluble (aqueous), no net reaction occurs.

You predict precipitates with solubility rules, which are very important to memorize. These rules are general AP-level shortcuts; a few exceptions exist, and exact solubility is handled later with $K_{sp}$ .

Usually soluble	Important exceptions
Group 1 cations and $\text{NH}_4^+$ salts	No common exceptions
$\text{NO}_3^-$ , $\text{C}_2\text{H}_3\text{O}_2^-$ / $\text{CH}_3\text{COO}^-$ , $\text{ClO}_3^-$ , $\text{ClO}_4^-$	No common exceptions
$\text{Cl}^-$ , $\text{Br}^-$ , $\text{I}^-$	Insoluble with $\text{Ag}^+$ , $\text{Pb}^{2+}$ , and $\text{Hg}_2^{2+}$
$\text{SO}_4^{2-}$	Insoluble or only slightly soluble with $\text{Ba}^{2+}$ , $\text{Sr}^{2+}$ , $\text{Pb}^{2+}$ , $\text{Ca}^{2+}$ , and $\text{Hg}_2^{2+}$

Usually insoluble	Important exceptions
$\text{CO}_3^{2-}$ , $\text{PO}_4^{3-}$ , $\text{CrO}_4^{2-}$ , $\text{S}^{2-}$	Soluble with Group 1 cations and $\text{NH}_4^+$
$\text{OH}^-$	Soluble with Group 1 cations and $\text{NH}_4^+$ ; $\text{Ba}^{2+}$ , $\text{Sr}^{2+}$ , and $\text{Ca}^{2+}$ hydroxides are more soluble than most other metal hydroxides

For a precipitation prediction, swap ion partners, apply the table, and write a net ionic equation only for the solid that forms. Example:

\text{Ag}^+(aq) + \text{Cl}^-(aq) \longrightarrow \text{AgCl}(s)

Example. Mix $20.0\ \mathrm{mL}$ of $0.100\ M$ silver nitrate with $30.0\ \mathrm{mL}$ of $0.100\ M$ sodium chloride. Assuming essentially complete precipitation, identify the excess reacting ion and its concentration.

There are $2.00\ \mathrm{mmol}$ silver ions and $3.00\ \mathrm{mmol}$ chloride. The one-to-one reaction leaves $1.00\ \mathrm{mmol}$ chloride in $50.0\ \mathrm{mL}$ , or $0.0200\ M$ . Sodium and nitrate remain spectators; they must still appear in a particle diagram of the final solution.

Acid–base (neutralization) reactions

Neutralization between a strong acid and strong base yields water and an ionic salt:

\text{acid} + \text{base} \longrightarrow \text{salt} + \text{H}_2\text{O}.

Such proton-transfer reactions are often rapid because water is a very stable product. This is a special form of a double-displacement reaction

Reactions of metal carbonates (and bicarbonates) with acid produce carbon dioxide as well, because carbonic acid is unstable and decomposes:

\text{H}_2\text{CO}_3 \longrightarrow \text{H}_2\text{O} + \text{CO}_2.

Metal oxides (basic anhydrides) react with acids like other bases as well, just giving up the oxygen to water instead of hydrogen.

Example. Equal volumes of $0.100\ M$ HCl and $0.100\ M$ barium hydroxide are mixed. Does equal molarity imply neutralization to equivalence?

Each mole of barium hydroxide supplies two moles of hydroxide. If each volume is $V$ liters, acid supplies $0.100V$ moles H+ while base supplies $0.200V$ moles OH-. The remaining $0.100V$ moles hydroxide occupy $2V$ liters, giving $0.0500\ M$ hydroxide. The mixture is basic.

Combination, decomposition, and combustion

Synthesis (combination) builds one product from simpler reactants, which can be elements or compounds:

\text{A} + \text{B} \longrightarrow \text{AB}.

Decomposition is the reverse picture: one compound breaks into two or more substances, often with heat or electricity. The products can be elements or compounds:

\text{AB} \longrightarrow \text{A} + \text{B}.

Combustion of a hydrocarbon in excess oxygen produces carbon dioxide and water (and other products if the fuel contains other elements such as sulfur). Incomplete combustion can yield $\text{CO}$ or soot (carbon):

\text{Hydrocarbon} + O_2 \longrightarrow CO_2 (g) + H_2O (g / l).

As usual, you should balance coefficients to make a balanced equation.

Example. Heating calcium carbonate produces calcium oxide and carbon dioxide. Classify the reaction and decide whether a decomposition reaction must produce elements.

The reaction is $\mathrm{CaCO_3\rightarrow CaO+CO_2}$ . One reactant gives multiple products, so this is decomposition, even though both products are compounds. Calcium stays $+2$ , carbon $+4$ , and oxygen $-2$ , so this decomposition is not redox.

Single-displacement reactions

In single-displacement, an element in its standard state replaces ions of another element in solution (or in a melt). For metals (and hydrogen in acid), activity order decides whether reaction occurs: a metal higher in the activity series reduces the cation of a metal below it. Hydrogen’s position marks which metals react with dilute acid to liberate $\text{H}_2$ .

More active metals	Tend to be oxidized more easily
$\mathrm{Li, K, Ba, Ca, Na}$	strongest reducing metals
$\mathrm{Mg, Al, Zn, Fe, Ni, Sn, Pb}$	can replace ions below them in solution
$\mathrm{H_2}$	reference: metals above can react with acids to form $\mathrm{H_2}$
$\mathrm{Cu, Ag, Pt, Au}$	least active; often remain unoxidized

For halogens, a more reactive halogen displaces the halide ion of a less reactive halogen from solution. Reactivity decreases down the group ( $\text{F}_2 > \text{Cl}_2 > \text{Br}_2 > \text{I}_2$ ).

Example. Metal X displaces copper from copper(II) solution but does not displace magnesium from magnesium(II) solution under comparable conditions. Place X in a relative activity series.

X is more readily oxidized than copper but less readily oxidized than magnesium. The supported order is $\mathrm{Mg>X>Cu}$ . These tests bracket X; they do not uniquely identify it or establish its position relative to every other metal.

Oxidation–reduction (redox)

Redox reactions transfer electrons between species. Oxidation is loss of electrons (increase in oxidation number); reduction is gain of electrons (decrease in oxidation number). A very helpful mnemonic is OIL RIG: Oxidation Is Losing, Reduction Is Gaining.

The reducing agent is the one being oxidized, and the oxidizing agent is the one being reduced. Assigning oxidation states to every atom in a formula is the standard bookkeeping method (refer to Unit 1). Many combustion, single-displacement, and electrochemical processes are redox; they are often slower in the lab than simple precipitation or strong acid–strong base neutralization because covalent bonds must break and form in the elemental or molecular reactants.

Assigning oxidation numbers (quick rules)

The whole framework depends on assigning oxidation numbers consistently. The standard priority list:

A free element is $0$ (e.g. $\text{O}_2$ , $\text{Na}$ , $\text{P}_4$ ).
A monatomic ion equals its charge ( $\text{Na}^+$ is $+1$ , $\text{S}^{2-}$ is $-2$ ).
Fluorine is always $-1$ in compounds; oxygen is usually $-2$ (but $-1$ in peroxides such as $\text{H}_2\text{O}_2$ and positive when bonded to F); hydrogen is $+1$ with nonmetals but $-1$ in metal hydrides such as $\text{NaH}$ .
The sum of oxidation numbers equals the overall charge of the species (zero for a neutral formula, the ion charge for a polyatomic ion).

Example. Assign oxygen’s oxidation number in $\mathrm{H_2O_2}$ and $\mathrm{OF_2}$ . Why does assuming oxygen is always $-2$ fail?

In peroxide, hydrogen is $+1$ , so $2(+1)+2x=0$ gives oxygen $-1$ . In $\mathrm{OF_2}$ , fluorine is $-1$ , so $x+2(-1)=0$ gives oxygen $+2$ . The sum must match the total charge, but the most reliable assignment rules must be applied first. Oxygen’s usual $-2$ value has specific exceptions.

Recognizing and classifying redox

If any element changes oxidation number from reactants to products, the reaction is redox. Combustion, single-displacement, and many synthesis/decomposition reactions are redox; most precipitation and acid–base reactions are not. A special case is disproportionation, where a single element is both oxidized and reduced in the same reaction—one portion goes up in oxidation number while another goes down (for example, $\text{Cl}_2$ reacting with cold base to give $\text{Cl}^-$ and $\text{ClO}^-$ ). The reverse, where two different oxidation states of an element combine into one intermediate state, is comproportionation.

Strong oxidizing agents (easily reduced) include $\text{F}_2$ , $\text{O}_2$ , $\text{Cl}_2$ , $\text{MnO}_4^-$ , $\text{Cr}_2\text{O}_7^{2-}$ , and concentrated $\text{HNO}_3$ . Strong reducing agents (easily oxidized) include reactive metals such as $\text{Na}$ , $\text{K}$ , and $\text{Mg}$ , as well as $\text{H}_2$ . These tendencies are quantified later as standard reduction potentials in Unit 9.

Example. Compare $\mathrm{HCl+NaOH\rightarrow NaCl+H_2O}$ with $\mathrm{2H_2+O_2\rightarrow2H_2O}$ . Both form water; why is only the second redox?

In neutralization, H remains $+1$ , O remains $-2$ , Na remains $+1$ , and Cl remains $-1$ . No oxidation numbers change. In the second reaction, hydrogen changes from $0$ to $+1$ and oxygen from $0$ to $-2$ , indicating oxidation and reduction. Product identity alone does not classify the electron-transfer behavior.

Half-reactions (oxidation and reduction)

A half-reaction shows only the oxidation or only the reduction part of an electron transfer. Electrons $(e^-)$ appear as a product in the oxidation half-reaction (electrons are lost) and as a reactant in the reduction half-reaction (electrons are gained). After each half-reaction is balanced for atoms and charge, you multiply one or both by integers so the number of electrons lost equals the number gained, then add the half-reactions and cancel duplicated species (including $e^-$ , $\text{H}_2\text{O}$ , $\text{H}^+$ , or $\text{OH}^-$ ).

Acidic solutions (common AP setup):

Basic solution: either balance as in acid and then add $\text{OH}^-$ to both sides in pairs that neutralize $\text{H}^+$ as water, or balance using $\text{H}_2\text{O}$ and $\text{OH}^-$ from the start. The final combined equation should contain no free $\text{H}^+$ if the medium is strongly basic.

Half-reactions describe oxidation at the anode and reduction at the cathode in galvanic and electrolytic cells. Tables of standard reduction potentials list half-reactions written as reduction by convention. You will learn more in Unit 9.

Example. In $2\mathrm{H_2O_2}\rightarrow2\mathrm{H_2O}+\mathrm{O_2}$ , identify the oxidized and reduced element and explain how one reactant fills both roles.

Oxygen begins at $-1$ in peroxide. Oxygen going to water becomes $-2$ and is reduced; oxygen going to oxygen gas becomes $0$ and is oxidized. Peroxide is therefore both oxidizing and reducing agent in this disproportionation reaction. Hydrogen remains $+1$ .

Solubility and “like dissolves like”

Solubility is the maximum amount of solute that dissolves in a given amount of solvent at a specified temperature. Polar and ionic solutes tend to dissolve in polar solvents (usually water); nonpolar solutes tend to dissolve in nonpolar solvents. At the particle level, ion–dipole interactions or hydrogen bonding stabilize ions or polar molecules in water; nonpolar solutes rely mainly on weaker dispersion forces to mix with nonpolar solvents.

By convention, the solvent is the component present in greater amount and the solute is the dissolved component (by convention in dilute lab work, the minor component).

Example. Ethanol dissolves in water but hexane barely does. Explain why saying ‘both contain carbon, so they should behave alike’ misses the relevant interactions.

Ethanol’s O-H group can form hydrogen bonds with water, compensating for some water-water interactions disrupted during mixing. Hexane cannot provide comparable polar interactions. The carbon skeleton is only part of the molecule; the functional group changes how it interacts with the solvent.

Factors affecting solubility

Common-ion effect: for a sparingly soluble salt in equilibrium with its solid, adding another source of the same cation or anion shifts equilibrium toward solid, lowering molar solubility. (The full equilibrium lecture is in Unit 7.)

Temperature: the effect on solid solubility depends on the sign of enthalpy of solution: many ionic solids become more soluble as $T$ rises, but exceptions exist. Gases in liquids almost always become less soluble as $T$ rises (molecular kinetic picture: escape from solution is easier when molecules move faster). Pressure has little effect on solid or liquid solutes but strongly affects gas solubility (Henry’s Law).

Example. A sealed carbonated drink is warmed and then opened. Separate the temperature effect from the pressure effect on dissolved carbon dioxide.

Warming generally lowers carbon dioxide solubility at a fixed gas pressure. Opening then lowers its partial pressure above the liquid, lowering the equilibrium dissolved concentration further. Gas escapes as the liquid approaches the new equilibrium; these are two distinct changes, not evidence that pressure and temperature are interchangeable.

Concentration measures (revisited)

Molarity $M$ is moles of solute per liter of solution ( $\text{mol/L}$ ); it changes slightly with temperature because volume changes.

Molality $m$ is moles of solute per kilogram of solvent (not kilogram of solution):

m = \frac{\text{moles of solute}}{\text{kilograms of solvent}}.

It is temperature-independent in the sense that it uses mass of solvent, not volume of solution.

Mass percent is

\text{mass \%} = \frac{\text{mass of solute}}{\text{mass of solution}} \times 100\%.

Parts per million and parts per billion report a mass ratio (or, for dilute aqueous work, often $\text{mg}$ solute per $\text{kg}$ solution, which is approximately $\text{mg/L}$ for water):

\text{ppm} = \frac{\text{mass of solute}}{\text{mass of sample}} \times 10^6, \qquad \text{ppb} = \frac{\text{mass of solute}}{\text{mass of sample}} \times 10^9.

Do not equate $\text{ppm}$ to “ $mass \% \times 10^6$ ”; use the fraction definition above (for example, $1\%$ corresponds to $10^4$ ppm).

Normality $N$ (still seen in some labs) is equivalents of reacting species per liter of solution: for acid–base, one equivalent of acid is one mole of $\text{H}^+$ donated per mole of formula (so $\text{H}_2\text{SO}_4$ can be $2N$ when both protons count in that context). On the AP exam, molarity and stoichiometry from the balanced equation are usually enough.

Example. A solution contains $10.0\ \mathrm{g}$ solute in $90.0\ \mathrm{g}$ water. It expands when warmed without evaporation. Determine its mass percent and which of molarity and molality changes.

Mass percent is $10.0/(10.0+90.0)\times100\%=10.0\%$ , not $11.1\%$ . Masses and mole amounts stay fixed, so molality stays fixed. Solution volume increases, so molarity decreases. The denominator determines which concentration measure responds to thermal expansion.

Colloids

A colloid contains dispersed particles larger than single molecules but small enough to stay suspended (roughly $1$ – $1000\ \text{nm}$ is a common textbook range). Colloids are thermodynamically unstable with respect to bulk phase separation but can be kinetically persistent; charged surfaces and electrostatic repulsion often slow aggregation.

The Tyndall effect is the scattering of visible light by those particles, a practical way to distinguish many colloids from true solutions. For example, milk is a colloid because it scatters light, one reason why it appears white
Coagulation (heating, adding electrolyte, or mixing) can collapse the dispersion so particles aggregate and settle, just like how milk clumps can settle when milk isn’t properly refrigerated

Example. Two apparently uniform samples transmit a flashlight beam, but only one shows a visible light path from the side. Explain what this supports and what it cannot prove.

A visible path indicates scattering by dispersed particles and supports a colloidal dispersion rather than a molecular solution. It does not establish the particles’ chemical identity or prove long-term stability. A macroscopically uniform appearance alone cannot distinguish a colloid from a true solution.

Practice

MCQ

Mix $25.0\ \mathrm{mL}$ of $0.200\ M$ barium chloride with $40.0\ \mathrm{mL}$ of $0.100\ M$ sodium sulfate. Assuming complete precipitation and additive volumes, what is the excess barium-ion concentration?

(A) $0.0615\ M$
(B) $0.0154\ M$
(C) $0.100\ M$
(D) $0.0769\ M$

For $2A+3B\rightarrow C$ , initially $0.40\ \mathrm{mol}$ A and $0.45\ \mathrm{mol}$ B react completely. Which result is correct?

(A) A limits; 0.20 mol C forms
(B) B limits; 0.30 mol C forms
(C) B limits; 0.15 mol C forms and 0.10 mol A remains
(D) A limits; 0.15 mol B remains

In acidic solution, permanganate is reduced to manganese(II) while iron(II) becomes iron(III). How many iron(II) ions react per permanganate ion?

(A) 1
(B) 2
(C) 3
(D) 5

A precipitate is weighed before it is fully dry and its mass is used to infer the original analyte concentration. Assuming no product loss, what bias results?

(A) Concentration is too high because retained water is counted as precipitate
(B) Concentration is too low because water has low molar mass
(C) No bias because water is neutral
(D) The sign cannot be predicted even with all assumptions stated

Which net ionic equation correctly represents neutralizing acetic acid with sodium hydroxide?

(A) $\mathrm{H^++OH^-\rightarrow H_2O}$ only
(B) $\mathrm{CH_3COOH+OH^-\rightarrow CH_3COO^-+H_2O}$
(C) $\mathrm{Na^++OH^-\rightarrow NaOH(s)}$
(D) $\mathrm{CH_3COO^-+H_2O\rightarrow CH_3COOH+OH^-}$

A hydrocarbon sample yields $0.30\ \mathrm{mol}$ carbon dioxide and $0.40\ \mathrm{mol}$ water upon complete combustion. Which empirical formula fits?

(A) $\mathrm{CH_2}$
(B) $\mathrm{C_3H_4}$
(C) $\mathrm{C_3H_8}$
(D) $\mathrm{C_6H_8}$

FRQ

A $2.50\ \text{g}$ sample of impure $\text{CaCO}_3$ reacts with excess $\text{HCl}$ according to

\text{CaCO}_3(s)+2\text{HCl}(aq)\rightarrow \text{CaCl}_2(aq)+\text{CO}_2(g)+\text{H}_2\text{O}(l).

The reaction produces $0.0200\ \text{mol}$ of $\text{CO}_2$ .

$(A)$ Calculate the moles of $\text{CaCO}_3$ that reacted.

$(B)$ Calculate the mass of $\text{CaCO}_3$ in the sample.

$(C)$ Calculate the percent by mass of $\text{CaCO}_3$ in the impure sample.

$(D)$ Original extension. Suppose $10.0\%$ of the carbon dioxide produced escapes before measurement. Recalculate the actual mass percent of calcium carbonate, assuming no other error.

A released AP Chemistry question asked students to identify a limiting reactant from experimental data. (Adapted from College Board, 2024 AP Chemistry FRQ 2.)

$(A)$ In a trial, $0.0300\ \text{mol}$ of $\text{Al}$ reacts with $0.0200\ \text{mol}$ of $\text{Cl}_2$ according to $2\text{Al}+3\text{Cl}_2\rightarrow2\text{AlCl}_3$ . Identify the limiting reactant.

$(B)$ Calculate the theoretical moles of $\text{AlCl}_3$ produced.

$(C)$ Explain why the excess reactant remains after the limiting reactant is consumed.

$(D)$ Original extension. In a second trial, double only the initial chlorine amount. Identify the limiting reactant and calculate the theoretical yield of aluminum chloride. Explain why doubling one reactant need not always double the yield.