Transition Elements · Practice

All questions with answers · Transition Elements

1. Bottle A has 2 g of salt in 100 mL of solution. Bottle B has 20 g of the same salt in 100 mL of solution. Which statement is correct?

   • D. They contain two different substances.
   • B. A is more concentrated: it has more water in it.
   • C. They are equally concentrated: both have the same volume.
   • A. B is more concentrated: it has more solute in the same volume.

   Answer: B is more concentrated: it has more solute in the same volume.

   Concentration compares the amount of solute in a given volume. Both bottles hold 100 mL. B holds 20 g of salt, and A only 2 g. So B is the concentrated one and A the dilute one. Both hold the same substance.

2. Solution P contains 12 g of a solute in 300 mL. Solution Q of the same solute has the same concentration, but its volume is only 50 mL. How many grams of solute does Q contain?

   Answer: 2 ± 0.01 g

   The same concentration means the same mass of solute in each mL. P: 12 g ÷ 300 mL = 0.04 g in each mL. Q: 0.04 g × 50 mL = 2 g. Most likely wrong answer: 12 g. A smaller volume of the same solution holds less solute. Another slip: 72 g, from multiplying instead of dividing.

3. An active metal reacts with a dilute acid. What are the products?

   • C. A metal oxide and hydrogen gas
   • D. A salt, water and carbon dioxide
   • A. A salt and hydrogen gas
   • B. A salt and water

   Answer: A salt and hydrogen gas

   Active metal + dilute acid → salt + hydrogen gas. The metal takes the place of the hydrogen in the acid. Example: Zn + 2HCl → ZnCl2 + H2$\mathrm{Zn}+2\mathrm{HCl}\to \mathrm{ZnCl}{}_2+\mathrm{H}{}_2$. A metal oxide gives a salt and water instead, with no gas.

4. Which equation is correct for aluminum with dilute sulphuric acid? (Al is +3, SO4$\mathrm{SO}{}_4$ is −2)

   • A. 2Al + 3H2SO4 → Al2(SO4)3 + 3H2$2\mathrm{Al}+3\mathrm{H}{}_2\mathrm{SO}{}_4\to \mathrm{Al}{}_2(\mathrm{SO}{}_4){}_3+3\mathrm{H}{}_2$
   • C. Al + H2SO4 → AlSO4 + H2$\mathrm{Al}+\mathrm{H}{}_2\mathrm{SO}{}_4\to \mathrm{AlSO}{}_4+\mathrm{H}{}_2$
   • D. 2Al + 3H2SO4 → Al2(SO4)3 + 6H$2\mathrm{Al}+3\mathrm{H}{}_2\mathrm{SO}{}_4\to \mathrm{Al}{}_2(\mathrm{SO}{}_4){}_3+6\mathrm{H}$
   • B. 2Al + 3H2SO4 → Al2(SO4)3 + 3H2O$2\mathrm{Al}+3\mathrm{H}{}_2\mathrm{SO}{}_4\to \mathrm{Al}{}_2(\mathrm{SO}{}_4){}_3+3\mathrm{H}{}_2\mathrm{O}$

   Answer: 2Al + 3H2SO4 → Al2(SO4)3 + 3H2$2\mathrm{Al}+3\mathrm{H}{}_2\mathrm{SO}{}_4\to \mathrm{Al}{}_2(\mathrm{SO}{}_4){}_3+3\mathrm{H}{}_2$

   Active metal + dilute acid → salt + hydrogen gas. Al is +3 and sulphate is −2, so the salt is Al2(SO4)3$\mathrm{Al}{}_2(\mathrm{SO}{}_4){}_3$. Two Al need three H2SO4$\mathrm{H}{}_2\mathrm{SO}{}_4$. Their six H atoms leave as three H2$\mathrm{H}{}_2$ molecules. Check: 2 Al, 6 H, 3 S and 12 O on each side.

5. Choose the correct answer: Pure iron is ...

   • C. relatively soft, malleable and ductile, but not magnetic
   • A. relatively soft, malleable, ductile and magnetic, and it melts at 1538 °C
   • D. very hard, so it is used in industry without adding other elements
   • B. very hard and brittle, magnetic, and it melts at 1538 °C

   Answer: relatively soft, malleable, ductile and magnetic, and it melts at 1538 °C

   The book describes pure iron as relatively soft, with low hardness, malleable, ductile and magnetic. It melts at 1538 °C, and its density is 7.87 g/cm³. Because it is relatively soft, pure iron has no industrial importance. Steels and iron alloys are used instead.

6. Write the balanced equation for the reaction of red hot iron with water vapour, with the condition over the arrow and state symbols.

   Answer: 3Fe(s) + 4H2O(v) →[500\,^C] Fe3O4(s) + 4H2(g)$3\mathrm{Fe}(\mathrm{s})+4\mathrm{H}{}_2\mathrm{O}(\mathrm{v})\stackrel{500{}_{}^{}{}_{}^{\circ }\mathrm{C}}{\to }\mathrm{Fe}{}_3\mathrm{O}{}_4(\mathrm{s})+4\mathrm{H}{}_2(\mathrm{g})$

   Red hot iron (500 °C) reacts with water vapour. The products are magnetic iron oxide and hydrogen gas. Fe3O4$\mathrm{Fe}{}_3\mathrm{O}{}_4$ needs 3 Fe and 4 O. Four H2O$\mathrm{H}{}_2\mathrm{O}$ supply the 4 O, and their 8 H atoms leave as four H2$\mathrm{H}{}_2$. Most likely wrong answers: Fe(OH)2$\mathrm{Fe}(\mathrm{OH}){}_2$ or Fe2O3$\mathrm{Fe}{}_2\mathrm{O}{}_3$ as the product. The book's product is Fe3O4$\mathrm{Fe}{}_3\mathrm{O}{}_4$.

7. Write the balanced equation for heating iron in chlorine gas, with state symbols.

   Answer: 2Fe(s) + 3Cl2(g) →[Δ] 2FeCl3(s)$2\mathrm{Fe}(\mathrm{s})+3\mathrm{Cl}{}_2(\mathrm{g})\stackrel{\mathrm{\Delta }}{\to }2\mathrm{FeCl}{}_3(\mathrm{s})$

   Iron reacts with chlorine to give iron (III) chloride, FeCl3$\mathrm{FeCl}{}_3$. Chlorine is a strong oxidizing agent, so iron reaches +3. Balance Cl: 3 Cl2$\mathrm{Cl}{}_2$ give 6 Cl, enough for 2 FeCl3$\mathrm{FeCl}{}_3$. Then write 2 Fe on the left. Most likely wrong answer: FeCl2$\mathrm{FeCl}{}_2$. That is the product with dilute HCl, not with chlorine gas.

8. Give reasons: Iron reacts with dilute mineral acids to give iron (II) salts, not iron (III) salts.

   Answer: Key points: (1) Iron displaces hydrogen from the dilute acid: Fe + 2HCl → FeCl2 + H2$\mathrm{Fe}+2\mathrm{HCl}\to \mathrm{FeCl}{}_2+\mathrm{H}{}_2$. (2) The hydrogen produced is a reducing agent. (3) It reduces iron (III) to iron (II), so iron (II) salts are obtained.

   Iron dissolves in dilute mineral acids, and hydrogen gas forms. The book's reason: iron (III) salts are reduced by the produced hydrogen. So the products are iron (II) salts, like FeSO4$\mathrm{FeSO}{}_4$ and FeCl2$\mathrm{FeCl}{}_2$. Key words: reduced by the produced hydrogen. Most likely wrong answer: 'iron (III) is unstable'. In fact iron (III), 3d⁵$3d^5$, is the more stable state.

9. Write the balanced equation for the reaction of iron with hot concentrated sulphuric acid.

   Answer: 3Fe(s) + 8H2SO4(l) →[conc.,\ Δ] FeSO4(aq) + Fe2(SO4)3(aq) + 4SO2(g) + 8H2O(v)$3\mathrm{Fe}(\mathrm{s})+8\mathrm{H}{}_2\mathrm{SO}{}_4(\mathrm{l})\stackrel{\text{conc.},\mathrm{\Delta }}{\to }\mathrm{FeSO}{}_4(\mathrm{aq})+\mathrm{Fe}{}_2(\mathrm{SO}{}_4){}_3(\mathrm{aq})+4\mathrm{SO}{}_2(\mathrm{g})+8\mathrm{H}{}_2\mathrm{O}(\mathrm{v})$

   Hot concentrated sulphuric acid gives iron (II) sulphate, iron (III) sulphate, sulphur dioxide and water. Check each side: 3 Fe, 8 S, 16 H and 32 O. Most likely wrong answer: Fe + H2SO4 → FeSO4 + H2$\mathrm{Fe}+\mathrm{H}{}_2\mathrm{SO}{}_4\to \mathrm{FeSO}{}_4+\mathrm{H}{}_2$, the dilute-acid equation. No hydrogen forms here.

10. Give reasons: Concentrated nitric acid causes the passivity of iron.

    Answer: Key points: (1) Concentrated nitric acid forms a thin layer of oxide on the surface of iron. (2) This layer protects the metal from further reaction. (3) The layer can be removed by abrasion or by dilute hydrochloric acid.

    Concentrated nitric acid does not dissolve iron. It forms a thin layer of oxide on the surface. The layer covers the iron and protects it from further reaction. This state is called passivity. Key words: thin layer of oxide, protects the metal from further reaction. Most likely wrong answer: 'iron does not react with acids'. Iron dissolves easily in dilute acids.

11. Which of the following statements about iron are correct? Choose all that apply.

    • S3. Many steels and iron alloys are produced with specific properties for many uses.
    • S4. Pure iron is not attracted by a magnet.
    • S6. Iron has a low melting point, so it is not used pure.
    • S1. Pure iron has no industrial importance because it is relatively soft.
    • S2. The physical properties of iron depend on its purity and the nature of its impurities.
    • S5. The purer the iron, the harder it is.

    Answer: Many steels and iron alloys are produced with specific properties for many uses.

    Pure iron is relatively soft, malleable, ductile and magnetic, and melts at 1538 °C. Being relatively soft, it has no industrial importance. Its physical properties depend on its purity and the nature of its impurities. So many steels and iron alloys are made for specific uses. Common slip: 'purer means harder'. Pure iron is the soft one.

12. Red hot iron is heated in dry oxygen. Which statement about the solid product is correct?

    • A. It is Fe3O4$\mathrm{Fe}{}_3\mathrm{O}{}_4$, a black oxide attracted by a magnet.
    • D. It is rust, a hydrated iron oxide.
    • C. It is FeO, because iron loses its two 4s electrons first.
    • B. It is Fe2O3$\mathrm{Fe}{}_2\mathrm{O}{}_3$, a red oxide like hematite.

    Answer: It is Fe3O4$\mathrm{Fe}{}_3\mathrm{O}{}_4$, a black oxide attracted by a magnet.

    Red hot iron reacts with dry air or oxygen to form magnetic iron oxide: 3Fe + 2O2 →[Δ] Fe3O4$3\mathrm{Fe}+2\mathrm{O}{}_2\stackrel{\mathrm{\Delta }}{\to }\mathrm{Fe}{}_3\mathrm{O}{}_4$. It is black and attracted by a magnet. It is the same compound as magnetite. Rust needs water, and the oxygen here is dry.

13. Iron filings are heated with sulphur powder. Which choice gives the product and the reason for iron's oxidation state in it?

    • A. FeS; sulphur is a weaker oxidizing agent than chlorine, so iron stops at +2.
    • B. Fe2S3$\mathrm{Fe}{}_2\mathrm{S}{}_3$; iron (III) has a stable half-filled 3d⁵$3d^5$ sublevel.
    • D. FeS; hydrogen formed in the reaction reduces iron (III) to iron (II).
    • C. FeSO4$\mathrm{FeSO}{}_4$; sulphur takes oxygen from the air during heating.

    Answer: FeS; sulphur is a weaker oxidizing agent than chlorine, so iron stops at +2.

    Iron and sulphur give iron (II) sulphide: Fe + S →[Δ] FeS$\mathrm{Fe}+\mathrm{S}\stackrel{\mathrm{\Delta }}{\to }\mathrm{FeS}$. Sulphur is −2, so iron is +2. Sulphur is a weaker oxidizing agent than chlorine, so iron stops at +2. No oxygen reacts, so no sulphate forms. No hydrogen forms either.

14. Write the balanced equation for the reaction of iron with dilute sulphuric acid, with state symbols.

    Answer: Fe(s) + H2SO4(aq) →[(dil)] FeSO4(aq) + H2(g)$\mathrm{Fe}(\mathrm{s})+\mathrm{H}{}_2\mathrm{SO}{}_4(\mathrm{aq})\stackrel{(\mathrm{dil})}{\to }\mathrm{FeSO}{}_4(\mathrm{aq})+\mathrm{H}{}_2(\mathrm{g})$

    Iron displaces hydrogen from the dilute acid. The salt is iron (II) sulphate, FeSO4$\mathrm{FeSO}{}_4$, and hydrogen gas is given off. The produced hydrogen keeps iron at +2. Most likely wrong answer: Fe2(SO4)3$\mathrm{Fe}{}_2(\mathrm{SO}{}_4){}_3$ as the salt.

15. Iron reacts with hot concentrated sulphuric acid. Which statement about the products is correct?

    • C. Iron (II) sulphate and hydrogen form, as with the dilute acid.
    • D. Iron (II) sulphate, iron (III) sulphate and hydrogen form.
    • B. Only iron (III) sulphate forms, because the concentrated acid is stronger.
    • A. Two salts form, with iron in the +2 and +3 oxidation states, and the gas is sulphur dioxide.

    Answer: Two salts form, with iron in the +2 and +3 oxidation states, and the gas is sulphur dioxide.

    The book's equation: 3Fe + 8H2SO4 → FeSO4 + Fe2(SO4)3 + 4SO2 + 8H2O$3\mathrm{Fe}+8\mathrm{H}{}_2\mathrm{SO}{}_4\to \mathrm{FeSO}{}_4+\mathrm{Fe}{}_2(\mathrm{SO}{}_4){}_3+4\mathrm{SO}{}_2+8\mathrm{H}{}_2\mathrm{O}$. Iron (II) sulphate holds iron at +2. Iron (III) sulphate holds iron at +3. The gas is sulphur dioxide, not hydrogen. The acid's sulphur is reduced from +6 to +4.

16. Write the scientific term: The state of iron when a thin layer of oxide, formed by concentrated nitric acid, protects it from further reaction.

    Answer: Passivity (of iron).

    Concentrated nitric acid forms a thin oxide layer on iron. The layer stops further reaction. This is passivity. It is not permanent: abrasion or dilute hydrochloric acid removes the layer. Common slip: 'rusting'. Rust needs water and air; the passive layer comes from concentrated nitric acid.

17. Sample	Hardness (made-up units)	On hammering
    A	60	bends slightly
    B	15	bends easily
    C	70	hardly bends

    The table shows tests on three iron samples. One of them is pure iron. Which sample is the pure iron, and why is it not used in industry?

    • C. B; it is not magnetic, so it has no use.
    • A. B; it is relatively soft, with low hardness.
    • D. C; it is too hard to be shaped.
    • B. C; pure iron is the hardest form of iron.

    Answer: B; it is relatively soft, with low hardness.

    Pure iron is relatively soft, with low hardness, so it bends easily. Sample B has the lowest hardness, so it is the pure iron. Samples A and C contain other elements, which make them harder. The book's reason: pure iron is relatively soft, so it has no industrial importance. Steels and iron alloys are used instead.

18. What happens when 16.8 g of iron is heated until red hot in excess dry oxygen? Write the balanced equation, then calculate the mass of the product. (Fe = 56, O = 16)

    Answer: Key points: (1) Magnetic iron oxide forms: 3Fe(s) + 2O2(g) →[Δ] Fe3O4(s)$3\mathrm{Fe}(\mathrm{s})+2\mathrm{O}{}_2(\mathrm{g})\stackrel{\mathrm{\Delta }}{\to }\mathrm{Fe}{}_3\mathrm{O}{}_4(\mathrm{s})$. (2) Moles of Fe = 16.8 ÷ 56 = 0.3 mol. (3) 3 mol Fe give 1 mol Fe3O4$\mathrm{Fe}{}_3\mathrm{O}{}_4$, so 0.1 mol forms. (4) Molar mass of Fe3O4$\mathrm{Fe}{}_3\mathrm{O}{}_4$ = 3(56) + 4(16) = 232 g/mol, so the mass = 0.1 × 232 = 23.2 g.

    Red hot iron and dry oxygen give magnetic iron oxide, Fe3O4$\mathrm{Fe}{}_3\mathrm{O}{}_4$. 16.8 g of iron is 0.3 mol. The equation shows 3 mol Fe → 1 mol Fe3O4$\mathrm{Fe}{}_3\mathrm{O}{}_4$. So 0.1 mol of Fe3O4$\mathrm{Fe}{}_3\mathrm{O}{}_4$ forms, which is 23.2 g. The solid gains mass because oxygen joins the iron. Most likely wrong answer: 24.0 g, from assuming Fe2O3$\mathrm{Fe}{}_2\mathrm{O}{}_3$ forms.

19. Compare between the reaction of iron with chlorine and its reaction with sulphur, in terms of: the equation, the oxidation state of iron in the product, and the reason for the difference.

    Answer: Key points: (1) Chlorine: 2Fe + 3Cl2 →[Δ] 2FeCl3$2\mathrm{Fe}+3\mathrm{Cl}{}_2\stackrel{\mathrm{\Delta }}{\to }2\mathrm{FeCl}{}_3$, iron (III) chloride, iron +3. (2) Sulphur: Fe + S →[Δ] FeS$\mathrm{Fe}+\mathrm{S}\stackrel{\mathrm{\Delta }}{\to }\mathrm{FeS}$, iron (II) sulphide, iron +2. (3) Chlorine is a stronger oxidizing agent than sulphur, so it oxidizes iron to +3; sulphur takes it only to +2.

    With chlorine, iron gives iron (III) chloride. Each Cl is −1, so iron is +3. With sulphur, iron gives iron (II) sulphide. Sulphur is −2, so iron is +2. Chlorine is the stronger oxidizing agent. It takes iron to +3, the stable half-filled 3d⁵$3d^5$ state. Most likely wrong answer: Fe2S3$\mathrm{Fe}{}_2\mathrm{S}{}_3$ for the sulphur reaction.

20. Iron filings are added to dilute hydrochloric acid, and the gas formed is collected. Which choice gives the salt in the solution and the gas collected?

    • A. FeCl2$\mathrm{FeCl}{}_2$; hydrogen
    • C. FeCl2$\mathrm{FeCl}{}_2$; chlorine
    • D. FeCl3$\mathrm{FeCl}{}_3$; chlorine
    • B. FeCl3$\mathrm{FeCl}{}_3$; hydrogen

    Answer: FeCl2$\mathrm{FeCl}{}_2$; hydrogen

    Iron is above hydrogen, so it displaces hydrogen from the acid: Fe + 2HCl →[(dil)] FeCl2 + H2$\mathrm{Fe}+2\mathrm{HCl}\stackrel{(\mathrm{dil})}{\to }\mathrm{FeCl}{}_2+\mathrm{H}{}_2$. The salt is iron (II) chloride. Iron (III) is reduced by the produced hydrogen. The chlorine stays in the salt; no chlorine gas forms.

21. Compare between the action of dilute sulphuric acid and of hot concentrated sulphuric acid on iron, in terms of: the iron salt(s) formed and the gas given off.

    Answer: Key points: (1) Dilute: Fe + H2SO4 → FeSO4 + H2$\mathrm{Fe}+\mathrm{H}{}_2\mathrm{SO}{}_4\to \mathrm{FeSO}{}_4+\mathrm{H}{}_2$; one salt, iron (II) sulphate; the gas is hydrogen. (2) Hot concentrated: 3Fe + 8H2SO4 → FeSO4 + Fe2(SO4)3 + 4SO2 + 8H2O$3\mathrm{Fe}+8\mathrm{H}{}_2\mathrm{SO}{}_4\to \mathrm{FeSO}{}_4+\mathrm{Fe}{}_2(\mathrm{SO}{}_4){}_3+4\mathrm{SO}{}_2+8\mathrm{H}{}_2\mathrm{O}$; two salts, iron (II) and iron (III) sulphates; the gas is sulphur dioxide (with water vapour).

    With the dilute acid, iron displaces hydrogen. The hydrogen keeps iron at +2, so only iron (II) sulphate forms. With the hot concentrated acid, no hydrogen forms. The products are iron (II) sulphate, iron (III) sulphate, sulphur dioxide and water. Most likely wrong answer: hydrogen gas from the concentrated acid too.

22. A passive iron sheet is put in dilute hydrochloric acid. After some time, bubbles of gas appear on it. Which explanation is correct?

    • C. The bubbles are oxygen, released as the oxide layer breaks down.
    • A. The acid removes the oxide layer; then fresh iron reacts with the acid and gives hydrogen.
    • B. The oxide layer itself reacts with the acid and gives hydrogen gas.
    • D. Passivity is permanent, so the gas must come from the acid breaking down by itself.

    Answer: The acid removes the oxide layer; then fresh iron reacts with the acid and gives hydrogen.

    Passivity is not permanent. Dilute hydrochloric acid removes the thin oxide layer. Fresh iron is then exposed. Iron is above hydrogen, so it reacts with the acid: Fe + 2HCl → FeCl2 + H2$\mathrm{Fe}+2\mathrm{HCl}\to \mathrm{FeCl}{}_2+\mathrm{H}{}_2$. The bubbles are hydrogen. Abrasion would also remove the layer.

23. Which reactants and condition give both magnetic iron oxide and hydrogen gas?

    • C. Iron and dilute hydrochloric acid at room temperature
    • D. Iron and liquid water at room temperature
    • A. Red hot iron (500 °C) and water vapour
    • B. Red hot iron and dry air

    Answer: Red hot iron (500 °C) and water vapour

    Red hot iron (500 °C) reacts with water vapour: 3Fe + 4H2O →[500\,^C] Fe3O4 + 4H2$3\mathrm{Fe}+4\mathrm{H}{}_2\mathrm{O}\stackrel{500{}_{}^{}{}_{}^{\circ }\mathrm{C}}{\to }\mathrm{Fe}{}_3\mathrm{O}{}_4+4\mathrm{H}{}_2$. Magnetic iron oxide and hydrogen gas form. With dry air, only Fe3O4$\mathrm{Fe}{}_3\mathrm{O}{}_4$ forms. With a dilute acid, an iron (II) salt and hydrogen form.

24. Metal X is the fourth most abundant element in the earth's crust, and its ore magnetite is black and magnetic. X is heated with sulphur. What is the product, and what is the oxidation state of X in it?

    • A. FeS; +2
    • B. Fe2S3$\mathrm{Fe}{}_2\mathrm{S}{}_3$; +3
    • D. Fe3S4$\mathrm{Fe}{}_3\mathrm{S}{}_4$; +2 and +3, like magnetic iron oxide
    • C. FeSO4$\mathrm{FeSO}{}_4$; +2

    Answer: FeS; +2

    Step 1: the fourth most abundant element in the crust, found in magnetite (Fe3O4$\mathrm{Fe}{}_3\mathrm{O}{}_4$), is iron. Step 2: iron and sulphur give iron (II) sulphide, Fe + S →[Δ] FeS$\mathrm{Fe}+\mathrm{S}\stackrel{\mathrm{\Delta }}{\to }\mathrm{FeS}$. Sulphur is −2, so iron is +2. Sulphur is a weaker oxidizing agent than chlorine, so iron does not reach +3.

25. Consider the following statements about the hydrogen produced when iron reacts with a dilute acid: (I) It acts as a reducing agent. (II) It reduces iron (III) to iron (II). (III) It oxidizes iron to iron (III). Which statements are correct?

    • D. (III) only
    • B. (I) and (III) only
    • A. (I) and (II) only
    • C. (II) and (III) only

    Answer: (I) and (II) only

    Hydrogen gas forms when iron displaces it from the acid. This hydrogen is a reducing agent. It reduces iron (III) to iron (II). So the products are iron (II) salts, not iron (III) salts. Statement (III) says the opposite of what happens.

26. 16.8 g of iron reacts completely with hot concentrated sulphuric acid: 3Fe + 8H2SO4 → FeSO4 + Fe2(SO4)3 + 4SO2 + 8H2O$3\mathrm{Fe}+8\mathrm{H}{}_2\mathrm{SO}{}_4\to \mathrm{FeSO}{}_4+\mathrm{Fe}{}_2(\mathrm{SO}{}_4){}_3+4\mathrm{SO}{}_2+8\mathrm{H}{}_2\mathrm{O}$. What mass of iron (III) sulphate forms? (Fe = 56, S = 32, O = 16)

    Answer: 40 ± 0.5 g

    16.8 g of iron is 16.8 ÷ 56 = 0.3 mol. The equation: 3 mol Fe give 1 mol Fe2(SO4)3$\mathrm{Fe}{}_2(\mathrm{SO}{}_4){}_3$. So 0.1 mol forms. Molar mass of Fe2(SO4)3$\mathrm{Fe}{}_2(\mathrm{SO}{}_4){}_3$ = 2(56) + 3(32 + 64) = 400 g/mol. Mass = 0.1 × 400 = 40 g. Most likely wrong answer: 120 g, from taking 0.3 mol of this salt. Only 2 of every 3 Fe atoms end in it.

27. Chromium is a highly reactive metal, yet it resists the atmosphere. Iron becomes passive in concentrated nitric acid. What do the two cases have in common?

    • B. Both metals lose their unpaired electrons and become unreactive.
    • D. In both, the protective layer is permanent and cannot be removed.
    • A. A thin oxide layer on the surface protects the metal from further reaction.
    • C. Both metals are below hydrogen, so air and acids cannot attack them.

    Answer: A thin oxide layer on the surface protects the metal from further reaction.

    Chromium forms a nonporous oxide layer on its surface. It prevents further reaction with the oxygen of air. Concentrated nitric acid forms a thin oxide layer on iron. It protects iron from further reaction. In both, a surface oxide layer protects the metal. On iron, the layer can be removed by abrasion or dilute hydrochloric acid.

28. What happens when 5.6 g of iron reacts completely with dilute hydrochloric acid? Write the balanced equation and calculate the mass of the salt formed. (Fe = 56, Cl = 35.5)

    Answer: Key points: (1) Fe(s) + 2HCl(aq) →[(dil)] FeCl2(aq) + H2(g)$\mathrm{Fe}(\mathrm{s})+2\mathrm{HCl}(\mathrm{aq})\stackrel{(\mathrm{dil})}{\to }\mathrm{FeCl}{}_2(\mathrm{aq})+\mathrm{H}{}_2(\mathrm{g})$: iron (II) chloride and hydrogen gas form. (2) Moles of Fe = 5.6 ÷ 56 = 0.1 mol. (3) 1 mol Fe gives 1 mol FeCl2$\mathrm{FeCl}{}_2$, so 0.1 mol forms. (4) Molar mass of FeCl2$\mathrm{FeCl}{}_2$ = 56 + 2(35.5) = 127 g/mol, so the mass = 12.7 g.

    Iron dissolves in dilute HCl. Hydrogen gas is given off, and iron (II) chloride forms. 5.6 g of iron is 0.1 mol, which gives 0.1 mol of FeCl2$\mathrm{FeCl}{}_2$. 0.1 × 127 = 12.7 g. Most likely wrong answer: 16.25 g, from writing FeCl3$\mathrm{FeCl}{}_3$ (molar mass 162.5 g/mol).

29. Match each ore in column (A) with the oxidation state of iron in it and its colour in column (B).

    Answer: Hematite (Fe2O3$\mathrm{Fe}{}_2\mathrm{O}{}_3$) → +3; blood red · Limonite (2Fe2O3.3H2O$2\mathrm{Fe}{}_2\mathrm{O}{}_3\cdot 3\mathrm{H}{}_2\mathrm{O}$) → +3; yellow · Siderite (FeCO3$\mathrm{FeCO}{}_3$) → +2; yellowish grey

    Hematite, Fe2O3$\mathrm{Fe}{}_2\mathrm{O}{}_3$: three O give −6, shared by two Fe, so iron is +3; blood red. Limonite, 2Fe2O3.3H2O$2\mathrm{Fe}{}_2\mathrm{O}{}_3\cdot 3\mathrm{H}{}_2\mathrm{O}$: the water adds no charge, so iron is again +3; yellow. Siderite, FeCO3$\mathrm{FeCO}{}_3$: carbonate is −2, so iron is +2; yellowish grey. Common slip: thinking the water in limonite changes iron's oxidation state.

30. Give reasons: A bar of pure iron bends more easily than a bar of steel of the same size.

    Answer: Key points: (1) In pure iron, atoms are arranged in compact closed rows, and a layer can slip over the other layers, so it is relatively soft. (2) Steel is an interstitial alloy: small carbon atoms enter the spaces of the iron lattice. (3) They prevent the movement of the layers, so steel is harder. Not accepted: "carbon replaces iron atoms" (that is a substitutional alloy).

    Pure iron is relatively soft. Its layers of atoms slip over each other when a force acts. In steel, carbon atoms with a small atomic size sit in the spaces of the iron lattice. They block the layers from slipping. So steel is harder and bends less. Key words: spaces of the crystal lattice, prevent the movement of the layers. Most likely wrong answer: "carbon atoms replace iron atoms in the lattice." That describes a substitutional alloy, where atoms of similar diameter take the places of the main metal's atoms. Carbon atoms are much smaller than iron atoms, so they do not take iron's places; they sit in the spaces between iron atoms, and every iron atom stays in place.

31. Consider the following statements (₂₆${}_{26}$Fe): (I) In FeCl3$\mathrm{FeCl}{}_3$, formed from iron and chlorine, iron has a half-filled 3d⁵$3d^5$ sublevel. (II) Iron gives iron (II) salts with dilute acids because Fe²⁺$\mathrm{Fe}{}^{2+}$ is more stable than Fe³⁺$\mathrm{Fe}{}^{3+}$. (III) The Fe²⁺$\mathrm{Fe}{}^{2+}$ ion has the configuration [Ar] 3d⁶$3d^6$. Which statements are correct?

    • A. (I) and (III) only
    • D. (I), (II) and (III)
    • B. (I) and (II) only
    • C. (II) and (III) only

    Answer: (I) and (III) only

    (I) is correct: in FeCl3$\mathrm{FeCl}{}_3$ iron is +3, and Fe³⁺$\mathrm{Fe}{}^{3+}$ is [Ar] 3d⁵$3d^5$, half-filled. (III) is correct: Fe loses its two 4s electrons to give [Ar] 3d⁶$3d^6$. (II) is wrong. Fe³⁺$\mathrm{Fe}{}^{3+}$ (3d⁵$3d^5$) is the more stable ion. Iron (II) salts form because the produced hydrogen reduces iron (III) to iron (II).