4.7 Types of Chemical Reactions

1. AIt is redox: an electron is transferred from the NH₃ molecule to the HCl molecule
   A student who thinks that the appearance of charges means electron transfer picks this. The charges arise because H⁺ moved; no oxidation number changes, so no electrons are transferred.
2. BIt is not acid-base: a hydroxide ion, OH⁻, is not one of the reactants in the reaction
   A student who thinks an acid-base reaction requires hydroxide ions picks this. An acid-base reaction is the transfer of a proton between species; here NH₃ accepts the proton.
3. CIt is not acid-base: a water molecule is not one of the products of the reaction
   A student who defines an acid-base reaction as one that gives a salt and water picks this. Water forms only when OH⁻ accepts the proton; the transfer of H⁺ from HCl to NH₃ is an acid-base reaction without it.
4. DIt is acid-base: a proton, H⁺, is transferred from HCl to the NH₃ molecule Correct
   The H that was attached to Cl is attached to N afterward, leaving Cl⁻ and forming NH₄⁺: a proton has been transferred between two species, which is what defines an acid-base reaction. The oxidation numbers (H +1, Cl −1, N −3) do not change.

CED 4.7.A.1 · Read this in Fix

1. AC₃H₈(g) + 3 O₂(g) → 3 CO₂(g) + 4 H₂(g)
   A student who thinks oxygen combines only with the carbon, with the hydrogen released as H₂, picks this. In complete combustion the hydrogen also combines with oxygen, forming H₂O.
2. BC₃H₈(g) + 10 O₂(g) → 3 CO₂(g) + 4 H₂O(g)
   A student who sets the coefficient of O₂ equal to the number of O atoms in the products (10) picks this. Each O₂ supplies two O atoms, so 5 O₂ are needed; this equation has 20 O atoms on the left and 10 on the right.
3. CC₃H₈(g) + 2 O₂(g) → 3 C(s) + 4 H₂O(g)
   A student who thinks burning leaves the carbon behind as a solid picks this. The atoms balance, but solid carbon forms only when combustion is incomplete; in complete combustion the carbon ends up as CO₂.
4. DC₃H₈(g) + 5 O₂(g) → 3 CO₂(g) + 4 H₂O(g) Correct
   The products of complete combustion of a hydrocarbon are CO₂ and H₂O. Three C atoms give 3 CO₂ and eight H atoms give 4 H₂O; these contain 10 O atoms, supplied by 5 O₂.

Working Complete combustion of a hydrocarbon: reactants are the hydrocarbon and O₂; products are CO₂ and H₂O. C: 3 → 3 CO₂. H: 8 → 4 H₂O. O atoms in products: 3(2) + 4(1) = 10, supplied by 5 O₂. C₃H₈(g) + 5 O₂(g) → 3 CO₂(g) + 4 H₂O(g).

CED 4.7.A.2 · Read this in Fix

1. AElectrons move from Cu²⁺ ions to the Zn atoms, so Zn is oxidized
   A student who thinks oxidation is the gain of electrons, and who sees from the charges that Zn is oxidized, concludes that Zn receives the electrons. A Zn atom becomes Zn²⁺ by losing two electrons, so the electrons move from Zn to Cu²⁺.
2. BElectrons move from the Zn atoms to Cu²⁺ ions, so Zn is reduced
   A student who takes 'reduced' to mean 'made smaller in amount' picks this, because the number of Zn atoms in the metal goes down. In chemistry the species that loses electrons is oxidized; the species reduced is Cu²⁺, which gains them.
3. CElectrons move from Cu²⁺ ions to Zn atoms, so Zn is reduced
   A student who thinks a positive ion carries extra electrons that it can pass on picks this. Cu²⁺ has two fewer electrons than a Cu atom; it gains electrons from Zn atoms, which become Zn²⁺ and are oxidized.
4. DElectrons move from Zn atoms to Cu²⁺ ions, so Zn is oxidized Correct
   Three Zn atoms become Zn²⁺ ions, so each has lost two electrons (oxidation number 0 to +2): Zn is oxidized. Three Cu²⁺ ions become Cu atoms, so each has gained two electrons. Electrons are transferred from the species oxidized, Zn, to the species reduced, Cu²⁺.

CED 4.7.A.3 · Read this in Fix

1. A+6
   A student who treats C₂ as a single particle stops at 2x = +6. The +6 is shared by two carbon atoms, so each is +3.
2. B+4
   A student who thinks oxidation numbers always add up to zero solves 2x + 4(−2) = 0. For an ion the total must equal the charge, 2−, which gives +3.
3. C+3 Correct
   Each O is −2 and the oxidation numbers add up to the ion's charge: 2x + 4(−2) = −2, so the two C atoms total +6 and each is +3.
4. D−2
   A student who thinks every atom in a polyatomic ion has the ion's charge as its oxidation number picks this. The 2− charge is the total of all six oxidation numbers, not the value for carbon.

Working Oxygen is −2. The oxidation numbers add up to the charge of the ion: 2x + 4(−2) = −2, so 2x = +6 and x = +3 for each carbon atom. Distractors: setting the sum to zero gives 2x = 8, x = +4; not dividing the +6 between the two C atoms gives +6; taking the ion's charge as the oxidation number gives −2.

CED 4.7.A.4 · Read this in Fix

1. AThe Ag⁺ and I⁻ ions have come together, keeping their charges, to form the solid Correct
   The solid at the bottom of the beaker is drawn as Ag⁺ and I⁻ ions packed together, each still carrying its charge. This is a precipitation reaction: ions combine to form an insoluble ionic compound, AgI(s), with no transfer of electrons.
2. BThe Ag⁺ ions have gained electrons from the I⁻ ions to form the solid
   A student who thinks ions must become neutral atoms to form a solid picks this. The diagram shows the particles in the solid still labeled Ag⁺ and I⁻, so no electrons have been transferred.
3. CThe K⁺ and NO₃⁻ ions have bonded together to form dissolved KNO₃ units
   A student who thinks the leftover ions join into formula units of the soluble product picks this. The diagram shows the K⁺ and NO₃⁻ ions far apart: they remain separate, unchanged ions in solution.
4. DThe K⁺ and NO₃⁻ ions have been used up along with the others in the solid
   A student who thinks all the ions are used up when a precipitate forms picks this. The diagram shows three K⁺ and three NO₃⁻ ions still in the solution; only Ag⁺ and I⁻ are in the solid.

CED 4.7.A.5 · Read this in Fix

1. AA precipitate forms, and it is potassium nitrate
   A student who thinks either new combination of ions can be the solid picks this. Mixture 3 contains K⁺ and NO₃⁻ ions together with no visible change, and all potassium and nitrate salts are soluble.
2. BA precipitate forms, and it is silver nitrate
   A student who thinks a precipitate is one of the original solutes coming back out of solution picks this. Silver nitrate is soluble: it is one of the starting solutions, and mixture 3 shows Ag⁺ and NO₃⁻ ions together with no solid.
3. CNo precipitate forms, and all four ions stay dissolved
   A student who thinks dissolved compounds stay as formula units, so that the chloride in KCl need not behave like the chloride in NaCl, picks this. The Cl⁻ ions from KCl are the same as those from NaCl, and mixture 1 shows that Ag⁺ and Cl⁻ form a solid.
4. DA precipitate forms, and it is silver chloride Correct
   Mixture 1 shows that a solid forms when Ag⁺ and Cl⁻ ions are in the same solution (the other combination there, NaNO₃, stays dissolved, as mixture 2 confirms). AgNO₃(aq) and KCl(aq) also bring Ag⁺ and Cl⁻ together, and mixture 3 shows that K⁺ and NO₃⁻ form no solid, so AgCl precipitates.

CED 4.7.A.5 · Read this in Fix

1. A1.60 mol
   A student who counts one oxygen atom for each product molecule picks this: 1.80 + 1.40 = 3.20 mol of O atoms, or 1.60 mol of O₂. Each CO₂ molecule contains two O atoms, so the products contain 2(1.80) + 1.40 = 5.00 mol of O atoms, which is 2.50 mol of O₂.
2. B2.50 mol Correct
   The hydrocarbon contains no oxygen, so all the O atoms in the products came from O₂: 2(1.80) + 1.40 = 5.00 mol of O atoms, which is 2.50 mol of O₂.
3. C3.20 mol
   A student who thinks each product molecule uses up one O₂ adds 1.80 mol and 1.40 mol. A molecule of H₂O contains only one O atom, so 1.40 mol of H₂O accounts for 0.70 mol of O₂.
4. D5.00 mol
   A student who thinks each O₂ supplies one oxygen atom takes the 5.00 mol of O atoms in the products as the amount of O₂. Each O₂ molecule contains two O atoms, so 2.50 mol reacted.

Working A hydrocarbon contains only C and H, so every O atom in the products came from O₂. Moles of O atoms = 2(1.80 mol) + 1(1.40 mol) = 5.00 mol. Moles of O₂ = 5.00 mol ÷ 2 = 2.50 mol. (The sample contained 1.80 mol C and 2.80 mol H.) Distractors: 5.00 mol is the moles of O atoms; 1.80 + 1.40 = 3.20 mol counts one O₂ per product molecule; (1.80 + 1.40) ÷ 2 = 1.60 mol counts one O atom per product molecule.

CED 4.7.A.2 · Read this in Fix

1. A1.0 mol
   A student who thinks each atom transfers one electron counts one electron for each of the 1.0 mol of Al atoms. Each Al atom goes from 0 to +3 and so transfers three electrons.
2. B2.0 mol
   A student who takes Fe as +2 in Fe₂O₃, the opposite of oxygen's −2, counts two electrons for each of the 1.0 mol of Fe atoms. The three O atoms total −6, so each Fe is +3 and gains three electrons.
3. C3.0 mol Correct
   Each Fe atom goes from +3 to 0, gaining three electrons, and 0.50 mol of Fe₂O₃ contains 1.0 mol of Fe atoms: 1.0 mol × 3 = 3.0 mol of electrons. The 1.0 mol of Al that reacts loses the same 3.0 mol.
4. D6.0 mol
   A student who adds the electrons lost by aluminum (3.0 mol) to the electrons gained by iron (3.0 mol) picks this. They are the same electrons, counted once.

Working Oxidation numbers: Fe is +3 in Fe₂O₃ and 0 in Fe, so each Fe gains 3 electrons; Al goes from 0 to +3, so each Al loses 3 electrons. 0.50 mol Fe₂O₃ contains 2 × 0.50 = 1.0 mol Fe, which gains 1.0 mol × 3 = 3.0 mol of electrons (equally, 1.0 mol Al loses 3.0 mol). Distractors: 1.0 mol × 2 = 2.0 mol takes Fe as +2; 3.0 + 3.0 = 6.0 mol adds electrons lost and gained; 1.0 mol assumes one electron per Al atom.

CED 4.7.A.3 · Read this in Fix

1. AThe claim is correct, because the oxidation number of zinc rises from 0 to +2 Correct
   Zn is 0 in the element and +2 in ZnCl₂, and H goes from +1 in HCl to 0 in H₂. Changes in oxidation numbers show that electrons are transferred (from Zn to H⁺), so the reaction is oxidation-reduction.
2. BThe claim is correct, because each zinc atom gains 2 electrons to form Zn²⁺
   A student who thinks an atom becomes a positive ion by gaining electrons picks this. The reaction is redox, but a Zn atom forms Zn²⁺ by losing two electrons, which are gained by H⁺ ions.
3. CThe claim is incorrect, because any reaction of an acid is acid-base instead
   A student who classifies every reaction of an acid as acid-base picks this. No proton is transferred to another species here: H⁺ ions gain electrons from Zn and form H₂, so the reaction is oxidation-reduction.
4. DThe claim is incorrect, because there is not a species that gains or loses oxygen
   A student who defines oxidation and reduction as the gain and loss of oxygen picks this. Redox reactions are transfers of electrons, shown by changes in oxidation numbers, and need not involve oxygen.

CED 4.7.A.4 · Read this in Fix

1. AEach Cu atom that reacted transferred one electron in all to Ag⁺
   A student who thinks each atom transfers one electron in a redox reaction picks this without using the data. The data show 0.0100 mol of Ag formed for 0.00500 mol of Cu, so each Cu atom supplies electrons to two Ag⁺ ions.
2. BEach Cu atom that reacted transferred three electrons in all to Ag⁺
   A student who takes the ratio of masses, 1.079 g ÷ 0.318 g ≈ 3.4, as the ratio of atoms picks this. Masses must be converted to moles: 0.0100 mol Ag to 0.00500 mol Cu is 2 to 1.
3. CEach Cu atom that reacted transferred two electrons in all to Ag⁺ Correct
   Cu reacted: 2.000 g − 1.682 g = 0.318 g, and 0.318 g ÷ 63.55 g/mol = 0.00500 mol. Ag formed: 1.079 g ÷ 107.87 g/mol = 0.0100 mol. Each Ag⁺ gains one electron to become Ag, and two Ag atoms form for each Cu atom that reacts, so each Cu atom transfers two electrons.
4. DEach Ag⁺ ion that reacted transferred one electron to the Cu atoms
   A student who thinks a positive ion carries extra electrons that it passes on picks this. Ag⁺ ions gain electrons to become Ag atoms; the electrons come from Cu atoms, which become Cu²⁺.

CED 4.7.A.3 · Read this in Fix

1. AEach Mg atom shares two electrons with an O atom, forming a separate molecule of MgO
   A student who thinks ionic compounds consist of molecules formed by sharing electrons picks this. Electrons are transferred from Mg to O, and MgO is a lattice of ions, not separate molecules.
2. BEach Mg atom transfers two electrons to an O atom, forming a lattice of Mg²⁺ and O²⁻ ions Correct
   Combustion is an oxidation-reduction reaction with O₂. Mg goes from 0 to +2 and O from 0 to −2: each Mg atom transfers two electrons to an O atom, and the resulting Mg²⁺ and O²⁻ ions make up the ionic solid MgO.
3. CEach O atom transfers two electrons to one Mg atom, forming a lattice of Mg²⁺ and O²⁻ ions
   A student who thinks an atom becomes a positive ion by gaining electrons picks this. Mg²⁺ has two fewer electrons than Mg; the electrons move from Mg to O.
4. DEach O₂ molecule stays unchanged and sticks to the Mg atoms, forming a powdery mixture
   A student who thinks burning just mixes or sticks the particles together unchanged picks this. MgO is a new substance: electrons are transferred, O₂ molecules are broken up, and Mg²⁺ and O²⁻ ions form.

CED 4.7.A.2 · Read this in Fix

1. AIt is inconsistent: four K⁺ ions should be drawn in the solution, not two Correct
   The two CO₃²⁻ ions in the solid came from two K₂CO₃ formula units, which also supply four K⁺ ions; the two Ca²⁺ ions came with four NO₃⁻ ions. The diagram shows four NO₃⁻ but only two K⁺, so atoms and charge are not conserved.
2. BIt is inconsistent: the K⁺ and NO₃⁻ ions should be drawn bonded as KNO₃ units
   A student who thinks the leftover ions join into formula units of the soluble product picks this. Potassium nitrate is soluble, so its ions are separate in solution, as drawn; the error is in the number of K⁺ ions.
3. CIt is inconsistent: the K⁺ and NO₃⁻ ions should all be drawn as part of the solid
   A student who thinks every ion is used up when a precipitate forms picks this. Potassium and nitrate salts are soluble, so K⁺ and NO₃⁻ stay in solution; only Ca²⁺ and CO₃²⁻ are in the solid.
4. DIt is consistent: two K₂CO₃ units supply two potassium ions and two CO₃²⁻ ions
   A student who treats K₂ in the formula as one particle picks this. Each K₂CO₃ formula unit supplies two separate K⁺ ions, so two units supply four, not the two drawn.

CED 4.7.A.5 · Read this in Fix

1. AIt increases, because oxygen atoms from the air become part of the solid Correct
   Combustion is a reaction with oxygen gas. Every iron atom stays on the pan, and oxygen atoms from O₂ in the air are added to the solid as iron oxide forms, so the reading increases.
2. BIt decreases, because part of the iron is destroyed as the steel wool burns
   A student who thinks burning destroys matter picks this. No atoms are destroyed: all the iron atoms remain in the solid oxide, and oxygen atoms are added.
3. CIt decreases, because burning gives off gases that then leave the balance pan
   A student who thinks every combustion gives off gases picks this. The only product of burning iron is a solid oxide, which stays on the pan; no gas is formed.
4. DIt stays the same, because mass is conserved in every chemical reaction
   A student who applies conservation of mass to the pan alone picks this. Mass is conserved for the iron and the oxygen together; the oxygen was not on the pan at the start, so the reading rises as it joins the solid.

CED 4.7.A.2 · Read this in Fix

1. A2 NO₃⁻(aq) + 2 NH₄⁺(aq) → 2 NH₄NO₃(s)
   A student who thinks either new combination of ions can be the solid picks this. All ammonium salts and all nitrate salts are soluble, so NH₄⁺ and NO₃⁻ remain in solution as spectator ions. The coefficients of 2 come from the full equation, 2 AgNO₃ + (NH₄)₂CrO₄ → Ag₂CrO₄ + 2 NH₄NO₃.
2. BCrO₄²⁻(aq) + Ag₂⁺(aq) → Ag₂CrO₄(s)
   A student who thinks the subscript stays with the ion writes the silver as one Ag₂⁺ particle. The subscript 2 in Ag₂CrO₄ counts two separate Ag⁺ ions, written 2 Ag⁺(aq); as written, the charge does not balance either.
3. CCrO₄²⁻(aq) + 2 Ag⁺(aq) → Ag₂CrO₄(s) Correct
   All ammonium and nitrate salts are soluble, so the precipitate cannot be NH₄NO₃ and must be the other new combination, Ag₂CrO₄. Two Ag⁺ ions balance the 2− charge of one CrO₄²⁻ ion; NH₄⁺ and NO₃⁻ are spectators.
4. DNO₃⁻(aq) + Ag⁺(aq) → AgNO₃(s)
   A student who thinks a precipitate is an original solute coming back out of solution picks this. AgNO₃ is one of the soluble starting compounds, and all nitrate salts are soluble; the precipitate is a new combination of ions from the two solutions.

Working Ions present: Ag⁺, NO₃⁻, NH₄⁺, CrO₄²⁻. The possible new combinations are NH₄NO₃ and Ag₂CrO₄. All ammonium and nitrate salts are soluble, so the precipitate is Ag₂CrO₄, and NH₄⁺ and NO₃⁻ are spectator ions. Net ionic equation: CrO₄²⁻(aq) + 2 Ag⁺(aq) → Ag₂CrO₄(s). Check: Ag 2 = 2, Cr 1 = 1, O 4 = 4; charge 2(1+) + (2−) = 0 = 0.

CED 4.7.A.5 · Read this in Fix

1. A0.10 mol
   A student who thinks every acid molecule transfers exactly one proton picks this. The equation shows H₄X becoming HX³⁻, a loss of three protons per molecule.
2. B0.30 mol Correct
   Each H₄X molecule ends up as HX³⁻, so it has transferred three protons, one to each of three OH⁻ ions: 0.10 mol × 3 = 0.30 mol.
3. C0.40 mol
   A student who thinks an acid always transfers all the hydrogen atoms in its formula multiplies by 4. In this reaction one H remains in HX³⁻, so only three protons per molecule are transferred.
4. D0.60 mol
   A student who thinks both H atoms of each water molecule formed came from the acid counts 6 protons per H₄X. Each OH⁻ already has one H and accepts one proton, so 3 H₂O represent 3 protons.

Working H₄X becomes HX³⁻, so each molecule transfers 3 protons (one to each of three OH⁻ ions, forming 3 H₂O). Moles of H⁺ transferred = 0.10 mol × 3 = 0.30 mol. Distractors: one proton per molecule gives 0.10 mol; all four H atoms give 0.40 mol; counting both H atoms of each of the 3 H₂O formed gives 0.10 × 6 = 0.60 mol.

CED 4.7.A.1 · Read this in Fix