Oxides Worked ExamplesIGCSE Chemistry 0620

First sort the list. Magnesium chloride and nitrogen are not oxides at all, so they can be ignored. That leaves four oxides: two of metals (copper, zinc) and two of non-metals (carbon, sulfur).

(a) Oxides of metals are generally basic. The syllabus names copper(II) oxide as a basic oxide, so the answer is copper(II) oxide. (Zinc oxide is an oxide of a metal too, but it is amphoteric — see (b).)

(b) An oxide that reacts with an acid and with a base to give a salt and water is amphoteric. Zinc oxide is one of the two amphoteric oxides in the syllabus, so the answer is zinc oxide.

(c) Oxides of non-metals are generally acidic: carbon dioxide and sulfur dioxide.

(d) The character of an oxide is related to the character of the element: copper is a metal, so its oxide is basic; sulfur is a non-metal, so its oxide is acidic.

(a) P is a metal — it is shiny and conducts electricity. Q is a non-metal — it is a dull yellow solid that does not conduct electricity.

(b) The oxide of a metal that dissolves in water gives an alkaline solution: universal indicator turns blue or purple, pH about $9$–$11$. The oxide of a non-metal that dissolves in water gives an acidic solution: universal indicator turns red or orange, pH about $2$–$4$.

(c) The oxide of P (a metal) is a basic oxide; the oxide of Q (a non-metal) is an acidic oxide. This links oxide character to metallic and non-metallic character.

(d) Sulfur combines with oxygen to form sulfur dioxide. Sulfur dioxide is toxic, so the experiment should be done in a fume cupboard (or a well-ventilated room).

Sort each formula first. Non-metal oxides ($\text{CO}_2$, $\text{SO}_2$) are acidic. $\text{CaO}$ and $\text{CuO}$ are basic. $\text{Al}_2\text{O}_3$ and $\text{ZnO}$ are amphoteric.

Row C — $\text{CO}_2$ acidic, $\text{CuO}$ basic, $\text{ZnO}$ amphoteric — matches all three. Answer: C.

A swaps the acidic and basic oxides (calcium is a metal, so $\text{CaO}$ is basic). B has $\text{SO}_2$ correct but puts amphoteric $\text{Al}_2\text{O}_3$ in the basic column and basic $\text{CuO}$ in the amphoteric column. D is tempting because $\text{CaO}$ is correctly basic, but zinc is a metal, so $\text{ZnO}$ cannot be acidic.

(a) Zinc oxide is amphoteric: an amphoteric oxide is one that reacts with acids and with bases to produce a salt and water.

(b) Acting as a base, zinc oxide gives the salt zinc chloride and water. Two $\text{HCl}$ are needed to supply the two chloride ions for $\text{ZnCl}_2$. Zinc oxide is a solid; the salt is aqueous; water is a liquid.

(c) Because this is also a reaction of an amphoteric oxide, the second product is water. The zincate has two Na, so $2\text{NaOH}$ is needed. Check: left Zn 1, O 3, Na 2, H 2; right Zn 1, Na 2, O $2 + 1 = 3$, H 2 ✓.

(d) The other amphoteric oxide named in the syllabus is aluminium oxide, $\text{Al}_2\text{O}_3$.

A solid element that does not conduct electricity is a non-metal (metals conduct). This rules out A and B.

A solution of pH $3$ is acidic, so the oxide is an acidic oxide — as expected for the oxide of a non-metal. Answer: D.

C pairs a non-metal with a basic oxide — a pH below $7$ cannot come from a basic oxide. B is wrong twice: E is not a metal, and amphoteric oxides (aluminium oxide, zinc oxide) do not dissolve in water to give an acidic solution.

An oxide that reacts with an acid but not with an alkali is basic; one that reacts with an alkali but not with an acid is acidic; one that reacts with both is amphoteric.

(a) W — basic (reacts with acid only). X — acidic (reacts with alkali only, and its solution has pH $2$). Y — amphoteric, because it reacts with both the acid and the alkali to give a salt and water. Z — basic (reacts with acid only, and its solution has pH $12$).

(b) Basic and amphoteric oxides are oxides of metals: W, Y and Z. X is acidic, so it is the oxide of a non-metal — consistent with it being a gas.

(c) W is a black, insoluble, basic oxide: copper(II) oxide, $\text{CuO}$.

(d) Z — it is basic and dissolves in water to form an alkaline solution (pH $12$), so it neutralises the acid in the soil. (Z behaves like calcium oxide.) W is also basic but is insoluble in water, so it would not spread through the soil; it is also not what farmers use.

(a) Observations are things you can see: the black solid dissolves (disappears) and the solution turns blue. (No gas is produced — an oxide, unlike a carbonate, does not fizz with acid.)

(b) A base + an acid gives a salt + water.

(c) The sulfate ions are spectator ions and are left out. Copper(II) oxide accepts protons ($\text{H}^+$ ions), so it is a base (bases are proton acceptors).

(d) Moles of acid: convert $\text{cm}^3$ to $\text{dm}^3$ first.

The mole ratio $\text{CuO} : \text{H}_2\text{SO}_4$ is $1:1$, so $0.0250$ mol of CuO is needed. $M_r(\text{CuO}) = 64 + 16 = 80$.

$0.0250$ mol of $\text{CuSO}_4$ forms. $M_r(\text{CuSO}_4) = 64 + 32 + 4 \times 16 = 160$.

(a) Nitric acid forms nitrates: aluminium nitrate. The aluminium ion is $\text{Al}^{3+}$ and the nitrate ion is $\text{NO}_3^{\,-}$, so three nitrate ions are needed: $\text{Al(NO}_3)_3$. The brackets are essential.

(b) One $\text{Al}_2\text{O}_3$ gives $2$ aluminium ions, needing $6$ nitrate ions, so $6\text{HNO}_3$; the $6$ H atoms form $3\text{H}_2\text{O}$. Check O: left $3 + 18 = 21$; right $2 \times 9 + 3 = 21$ ✓.

(c) $M_r(\text{Al}_2\text{O}_3) = 2 \times 27 + 3 \times 16 = 102$.

The ratio $\text{Al}_2\text{O}_3 : \text{HNO}_3$ is $1:6$.

(d) The ratio $\text{Al}_2\text{O}_3 : \text{Al(NO}_3)_3$ is $1:2$, so $0.100$ mol of salt. $M_r = 27 + 3 \times (14 + 3 \times 16) = 213$.

Percentage yield = actual ÷ theoretical × 100.

(a) Before: $84.0\text{ cm}^3$. After: $60.0\text{ cm}^3$. Only carbon dioxide is absorbed by the alkali, so the decrease is the volume of carbon dioxide.

(b) Percentage by volume:

(c) Carbon dioxide reacts with the alkali to form a salt and water, which shows it is an acidic oxide.

(d) Convert to $\text{dm}^3$ and use the molar gas volume, $24\text{ dm}^3$ at r.t.p.

The ratio $\text{CO}_2 : \text{Na}_2\text{CO}_3$ is $1:1$. $M_r(\text{Na}_2\text{CO}_3) = 2 \times 23 + 12 + 3 \times 16 = 106$.

(e) So that all of the carbon dioxide has reacted; a constant volume shows no more is being absorbed, so the decrease is the full volume of carbon dioxide.

(a) From left to right the pH falls from alkaline ($13$, $10$) to acidic ($2$, $3$): the oxides change from basic (sodium, magnesium) through amphoteric (aluminium) to acidic (silicon, phosphorus, sulfur).

(b) Across a period the elements change from metals to non-metals. Metal oxides are basic and non-metal oxides are acidic, so the oxides follow the same trend.

(c) A pH of $7$ only shows that the oxide does not dissolve in water — it cannot change the pH. To classify an insoluble oxide you must test whether it reacts with an acid, an alkali, or both.

Aluminium oxide reacts with acids and with alkalis, so it is amphoteric. Silicon dioxide is the oxide of a non-metal and is acidic — for example it reacts with the basic oxide calcium oxide to form calcium silicate (slag) in the blast furnace.

(d) A soluble basic oxide reacts with water to form an alkali.

(a) Q reacts with the acid (dilute nitric acid) and with the alkali (warm sodium hydroxide) to form salts. So Q is amphoteric.

(b) A white precipitate that dissolves in excess sodium hydroxide could be $\text{Al}^{3+}$ or $\text{Zn}^{2+}$. With aqueous ammonia, $\text{Al}^{3+}$ gives a white precipitate that is insoluble in excess, but $\text{Zn}^{2+}$ gives a white precipitate that dissolves in excess. The precipitate dissolved, so the ion is zinc, $\text{Zn}^{2+}$.

(c) Q is zinc oxide, one of the two amphoteric oxides. With nitric acid it forms zinc nitrate: $\text{Zn}^{2+}$ needs two $\text{NO}_3^{\,-}$ ions, so the formula is $\text{Zn(NO}_3)_2$.

(d) The white precipitate is zinc hydroxide. Charges balance: $+2$ and $2 \times (-1)$ give a neutral solid.

(a) Aluminium oxide is amphoteric, so it reacts with the alkali and passes through the filter paper in solution (filtrate T). Iron(III) oxide is a basic oxide, so it does not react with the alkali and stays behind: residue S is iron(III) oxide.

(b) $\text{Al}_2\text{O}_3$ contains two Al, so $2\text{NaAlO}_2$ forms, needing $2\text{NaOH}$; the two H atoms form one water molecule. Check O: $3 + 2 = 5$ on the left, $4 + 1 = 5$ on the right ✓.

(c) The rest of the sample was aluminium oxide.

(d) $M_r(\text{Al}_2\text{O}_3) = 102$.

The ratio $\text{Al}_2\text{O}_3 : \text{NaOH}$ is $1:2$.

(e) Solution containing dissolved sodium aluminate and excess sodium hydroxide is left on the residue. If it is not washed off, these solids remain when it dries and the mass of S would be too high, so the percentage of aluminium oxide would be too low.