Chemistry of Non-metals

Why HBr cannot be made this way: conc. H$_2$SO$_4$ is an oxidising agent. HBr is a strong reducing agent, so as soon as it forms it is oxidised to bromine:

$\mathrm{NaBr + H_2SO_4 \rightarrow NaHSO_4 + HBr}$ $\mathrm{2HBr + H_2SO_4 \rightarrow Br_2 + SO_2 + 2H_2O}$

So the product is contaminated with Br$_2$ and SO$_2$ instead of pure HBr (HBr is made using non-oxidising conc. H$_3$PO$_4$ instead).

Preparation of HCl: heating sodium chloride with conc. H$_2$SO$_4$ (below 200 °C):

$\mathrm{NaCl + H_2SO_4 \xrightarrow{\Delta} NaHSO_4 + HCl}$

Test for the ions: add AgNO$_3$ solution, then dilute HNO$_3$, then ammonia solution.

• Cl$^-$: white curdy precipitate of AgCl, readily soluble in dilute NH$_4$OH.
• Br$^-$: pale yellow precipitate of AgBr, only sparingly soluble in dilute NH$_4$OH.

$\mathrm{Ag^+ + Cl^- \rightarrow AgCl\downarrow}\qquad \mathrm{Ag^+ + Br^- \rightarrow AgBr\downarrow}$

Why HI cannot be prepared: HI is a strong reducing agent. Concentrated $\mathrm{H_2SO_4}$ is an oxidising agent, so the HI formed is oxidised to iodine, and the acid is reduced to $\mathrm{SO_2}$ (or $\mathrm{H_2S}$):

$\mathrm{2HI + H_2SO_4 \rightarrow I_2 + SO_2 + 2H_2O}$

Hence HI is prepared with non-oxidising acids such as concentrated $\mathrm{H_3PO_4}$.

Preparation of HCl:

$\mathrm{NaCl + H_2SO_4 \xrightarrow{\ \ \ } NaHSO_4 + HCl\uparrow}$

Tests:

• For $\mathrm{Cl^-}$: add $\mathrm{AgNO_3}$ solution. A white precipitate of AgCl forms which dissolves in $\mathrm{NH_4OH}$: $\mathrm{Cl^- + Ag^+ \rightarrow AgCl\downarrow}$
• For $\mathrm{Br^-}$: $\mathrm{AgNO_3}$ gives a pale yellow precipitate of AgBr, which is only sparingly soluble in $\mathrm{NH_4OH}$.

Tritium ($^3_1$H or T) is the least abundant isotope of hydrogen (about 1 atom in $10^{17}$ hydrogen atoms; it is radioactive).

Uses: (1) as a radioactive tracer to study reaction mechanisms and biological processes; (2) as fuel in nuclear fusion reactions and hydrogen bombs.

Ozone depletion is mainly caused by chlorofluorocarbons (CFCs) and similar chemicals that release chlorine radicals in the stratosphere. It can be reduced by:

• stopping the use and production of CFCs, halons and carbon tetrachloride, and replacing them with ozone-friendly substitutes (HFCs, hydrocarbons) in refrigerators, air-conditioners and aerosol sprays;
• recovering and recycling refrigerant gases from old appliances instead of releasing them;
• limiting emissions of nitrogen oxides from high-flying aircraft and following international agreements such as the Montreal Protocol.

Laboratory preparation: dehydration of formic (methanoic) acid with warm conc. H$_2$SO$_4$:

$\mathrm{HCOOH \xrightarrow[\Delta]{conc.\ H_2SO_4} CO + H_2O}$

CO is harmful because it is a poisonous gas that combines with haemoglobin in blood about 200-300 times more strongly than oxygen does, forming stable carboxyhaemoglobin. This stops blood carrying oxygen to tissues and can cause suffocation and death.

(i) Boric acid: H$_3$BO$_3$. Use: as a mild antiseptic and eye wash.

(ii) Orthophosphoric acid: H$_3$PO$_4$. Use: in the manufacture of phosphate fertilizers (it is also used in soft drinks and rust removers).

Kipp's apparatus is a glass apparatus of three vertically joined parts: a top globe with a funnel, a middle bulb and a bottom bulb.

Method

1. Lumps of iron(II) sulphide (FeS) are placed in the middle bulb (through the side opening), which has a gas outlet with a stopcock.
2. Dilute HCl (or dilute H$_2$SO$_4$) is poured through the top funnel. It goes to the bottom bulb and rises up to the FeS in the middle bulb.
3. When the stopcock is opened, the acid reacts with FeS and H$_2$S gas is evolved:

$\mathrm{FeS + 2HCl \rightarrow FeCl_2 + H_2S\uparrow}$

4. When the stopcock is closed, the gas pressure pushes the acid back into the top globe, the acid leaves the FeS, and the reaction stops automatically. Opening the stopcock restarts it.
5. The gas is passed through water to remove HCl fumes, dried with anhydrous CaCl$_2$ (not conc. H$_2$SO$_4$, which oxidises H$_2$S), and collected by downward displacement of air (it is heavier than air).

H$_2$S as a reducing agent: it reduces chlorine to HCl, being itself oxidised to sulphur:

$\mathrm{H_2S + Cl_2 \rightarrow 2HCl + S\downarrow}$

Para hydrogen is more stable. In para hydrogen the two nuclear spins are antiparallel, giving a lower energy state. In ortho hydrogen the spins are parallel, which has higher energy. At very low temperature (near 0 K) hydrogen is almost entirely para.

Chlorofluorocarbons (e.g. $\mathrm{CF_2Cl_2}$) are stable in the lower atmosphere and drift into the stratosphere. There, ultraviolet radiation breaks the C-Cl bond and releases chlorine free radicals:

$\mathrm{CF_2Cl_2 \xrightarrow{UV} \cdot CF_2Cl + Cl\cdot}$

The chlorine radical destroys ozone in a chain reaction:

$\mathrm{Cl\cdot + O_3 \rightarrow ClO\cdot + O_2}$ $\mathrm{ClO\cdot + O \rightarrow Cl\cdot + O_2}$

The chlorine radical is regenerated, so one Cl atom can destroy thousands of ozone molecules, thinning the ozone layer.

Allotropy is the property by which an element exists in two or more different physical forms in the same state, having the same chemical properties but different physical properties (e.g. diamond and graphite for carbon).

The hardest allotrope of carbon is diamond. It is used for cutting glass and drilling rocks, and as a gemstone in jewellery.

Preparation: Phosphine is prepared by heating white phosphorus with concentrated NaOH solution in an inert atmosphere:

$\mathrm{P_4 + 3NaOH + 3H_2O \rightarrow PH_3\uparrow + 3NaH_2PO_2}$

With $\mathrm{O_2}$: Phosphine burns in air or oxygen to form phosphorus pentoxide (or phosphoric acid) and water:

$\mathrm{2PH_3 + 4O_2 \rightarrow P_2O_5 + 3H_2O}$

• $\mathrm{BaO_2}$: peroxide (contains the $\mathrm{O_2^{2-}}$ ion)
• $\mathrm{Al_2O_3}$: amphoteric oxide (reacts with both acids and bases)
• $\mathrm{Fe_3O_4}$: mixed (compound) oxide ($\mathrm{FeO\cdot Fe_2O_3}$)
• $\mathrm{N_2O_3}$: acidic oxide (gives HNO₂ with water)

• (i) Ozone layer: absorbs harmful ultraviolet radiation from the sun, protecting life on Earth.
• (ii) Heavy water (D₂O): used as a moderator to slow neutrons in nuclear reactors.
• (iii) Dihydrogen: used to make ammonia by the Haber process (also as a rocket fuel and for hydrogenation of oils).
• (iv) Tritium: a radioactive isotope used as a tracer and as fuel in fusion (hydrogen bomb).

Preparation: oxalic acid is heated with conc. H₂SO₄, which dehydrates it. $\mathrm{(COOH)_2 \xrightarrow{conc.\ H_2SO_4,\ \Delta} CO + CO_2 + H_2O}$ (CO₂ is removed by passing the gas through NaOH solution.)

With hot nickel: CO combines with finely divided nickel at about 330-350 K to form volatile nickel tetracarbonyl. $\mathrm{Ni + 4CO \xrightarrow{330-350\ K} Ni(CO)_4}$ On stronger heating (about 450 K) it decomposes back to pure Ni (Mond process for refining nickel).

Responsible chemicals: oxides of sulphur and nitrogen ($\mathrm{SO_2}$, $\mathrm{NO_x}$), which react with water vapour and oxygen in the air to form H₂SO₄ and HNO₃ (CO₂ adds a little carbonic acid). $\mathrm{2SO_2 + O_2 + 2H_2O \rightarrow 2H_2SO_4}$

Harmful effects: it corrodes buildings, marble monuments and metals; acidifies lakes and soil, killing fish and damaging crops and forests.

Laboratory preparation: heating MnO₂ with conc. HCl. $\mathrm{MnO_2 + 4HCl \xrightarrow{\Delta} MnCl_2 + Cl_2 + 2H_2O}$

Actions of Cl₂

• (i) NH₃: with excess ammonia, nitrogen is formed; with excess chlorine, explosive nitrogen trichloride forms. $\mathrm{8NH_3 + 3Cl_2 \rightarrow 6NH_4Cl + N_2} \qquad \mathrm{NH_3 + 3Cl_2 \rightarrow NCl_3 + 3HCl}$
• (ii) Conc. NaOH (hot): forms chloride and chlorate. $\mathrm{6NaOH + 3Cl_2 \rightarrow 5NaCl + NaClO_3 + 3H_2O}$ (with cold dilute NaOH, NaClO is formed instead).
• (iii) H₂S: chlorine oxidises it, and yellow sulphur is deposited. $\mathrm{H_2S + Cl_2 \rightarrow 2HCl + S}$

Test for Cl⁻: add dilute HNO₃ and then AgNO₃ solution to the solution. A white curdy precipitate of AgCl forms, which dissolves in ammonia solution. $\mathrm{Ag^+ + Cl^- \rightarrow AgCl\downarrow} \qquad \mathrm{AgCl + 2NH_3 \rightarrow [Ag(NH_3)_2]Cl}$

Ostwald process

Principle: ammonia is catalytically oxidised to nitric oxide, which is further oxidised and absorbed in water.

1. Catalytic oxidation of NH₃: a mixture of one part ammonia and about ten parts air is passed over a red-hot platinum-rhodium gauze at about 800-900 °C (and 4-5 atm). The reaction is exothermic, so the catalyst stays hot without extra heating. $\mathrm{4NH_3 + 5O_2 \xrightarrow{Pt\text{-}Rh,\ 800\,^\circ C} 4NO + 6H_2O}$
2. Oxidation of NO: the gas is cooled and mixed with more air; NO is oxidised to brown NO₂. $\mathrm{2NO + O_2 \rightarrow 2NO_2}$
3. Absorption in water: NO₂ is passed with air into water in absorption towers. $\mathrm{4NO_2 + O_2 + 2H_2O \rightarrow 4HNO_3}$ This gives about 60-68% HNO₃, which can be concentrated by distillation to about 68% (the azeotrope), then dehydrated using conc. H₂SO₄ to get higher strength.

Overall: $\mathrm{NH_3 + 2O_2 \rightarrow HNO_3 + H_2O}$.

Laboratory preparation of $\mathrm{H_2S}$. Hydrogen sulphide is prepared in Kipp's apparatus by the action of dilute sulphuric (or hydrochloric) acid on ferrous sulphide:

$\mathrm{FeS + H_2SO_4(dil) \rightarrow FeSO_4 + H_2S\uparrow}$

The gas is passed through water to remove acid spray and dried with phosphorus pentoxide, $\mathrm{P_4O_{10}}$ (conc. $\mathrm{H_2SO_4}$ and $\mathrm{CaO}$ cannot be used because they react with $\mathrm{H_2S}$). It is heavier than air, so it is collected by upward delivery (downward displacement of air). It has a rotten-egg smell and is poisonous, so the preparation is done in a fume cupboard.

Action of $\mathrm{H_2S}$ (a reducing agent).

(i) With acidified $\mathrm{K_2Cr_2O_7}$ the orange solution turns green as $\mathrm{Cr^{6+}}$ is reduced to $\mathrm{Cr^{3+}}$, and yellow sulphur is deposited:

$\mathrm{K_2Cr_2O_7 + 4H_2SO_4 + 3H_2S \rightarrow K_2SO_4 + Cr_2(SO_4)_3 + 7H_2O + 3S\downarrow}$

(ii) With $\mathrm{FeCl_3}$ the yellow-brown solution turns pale green as $\mathrm{Fe^{3+}}$ is reduced to $\mathrm{Fe^{2+}}$, and sulphur is precipitated:

$\mathrm{2FeCl_3 + H_2S \rightarrow 2FeCl_2 + 2HCl + S\downarrow}$

Application in salt analysis. $\mathrm{H_2S}$ is used as a group reagent for basic radicals (cations). In presence of dilute HCl (low $\mathrm{S^{2-}}$ concentration) it precipitates the Group II sulphides: $\mathrm{PbS}$ (black), $\mathrm{CuS}$ (black), $\mathrm{CdS}$ (yellow), $\mathrm{Bi_2S_3}$ (brown), $\mathrm{As_2S_3}$ (yellow). In presence of $\mathrm{NH_4Cl}$ and $\mathrm{NH_4OH}$ it precipitates Group IV sulphides: $\mathrm{ZnS}$ (white), $\mathrm{MnS}$ (buff/pink), $\mathrm{NiS}$ and $\mathrm{CoS}$ (black). The colour of the sulphide helps to identify the cation. Paper soaked in lead acetate turning black is also used to detect $\mathrm{H_2S}$ (sulphide radical).

(i) $\mathrm{Al_2O_3}$ reacts with both acids and bases to give salt and water, so it is amphoteric:

$\mathrm{Al_2O_3 + 6HCl \rightarrow 2AlCl_3 + 3H_2O} \qquad \mathrm{Al_2O_3 + 2NaOH \rightarrow 2NaAlO_2 + H_2O}$

(ii) $\mathrm{BaO_2}$ contains the peroxide ion $\mathrm{O_2^{2-}}$ (with an $\mathrm{O-O}$ link), and it gives hydrogen peroxide with dilute acids:

$\mathrm{BaO_2 + H_2SO_4 \rightarrow BaSO_4 + H_2O_2}$

Para-hydrogen is the form of molecular hydrogen in which the nuclear spins of the two protons are in opposite (antiparallel) directions. It is the low-energy form and is the only form present near $0\ \text{K}$.

Conversion to ortho-hydrogen (parallel spins) is brought about by:

• heating to high temperature (above about $800^\circ\mathrm{C}$) or passing an electric discharge,
• passing it over a catalyst such as platinum or iron,
• contact with atomic hydrogen, or with paramagnetic substances such as $\mathrm{O_2}$ or $\mathrm{NO}$.

At room temperature, ordinary hydrogen is an equilibrium mixture of about $75\%$ ortho and $25\%$ para.

(i) When CO is heated with $\mathrm{H_2}$ in presence of $\mathrm{ZnO}$ and $\mathrm{Cu}$ catalysts (about $573\ \text{K}$, $200$ to $300$ atm), methanol is formed:

$\mathrm{CO + 2H_2 \xrightarrow{ZnO + Cu} CH_3OH}$

(ii) When CO is passed over heated caustic soda under pressure (about $200^\circ\mathrm{C}$, $10$ atm), sodium formate is formed:

$\mathrm{CO + NaOH \xrightarrow{\Delta,\ pressure} HCOONa}$

(i) Boric acid: $\mathrm{H_3BO_3}$. Use: mild antiseptic and eye wash.

(ii) Borax: $\mathrm{Na_2B_4O_7\cdot 10H_2O}$. Use: in the borax-bead test for metal ions, and in making borosilicate (Pyrex) glass.

(iii) Silica gel: $\mathrm{SiO_2\cdot xH_2O}$. Use: drying agent (dehumidifier) and adsorbent in chromatography.

(iv) Sodium silicate: $\mathrm{Na_2SiO_3}$. Use: in detergents and soaps, and as an adhesive and fire-proofing agent (water glass).

Why HBr and HI cannot be prepared with conc. $\mathrm{H_2SO_4}$

Conc. $\mathrm{H_2SO_4}$ is an oxidising agent, while HBr and HI are strong reducing agents (HI more than HBr). The HBr or HI first formed is therefore oxidised by the acid:

$\mathrm{2HBr + H_2SO_4 \rightarrow Br_2 + SO_2 + 2H_2O}$

$\mathrm{2HI + H_2SO_4 \rightarrow I_2 + SO_2 + 2H_2O}$

So the products are $\mathrm{Br_2}$ or $\mathrm{I_2}$ (with $\mathrm{SO_2}$), and the pure halogen acid is not obtained. (Chloride is not oxidised this way, so HCl can be made from NaCl and conc. $\mathrm{H_2SO_4}$.)

Correct preparation

Use a non-oxidising acid such as conc. $\mathrm{H_3PO_4}$:

$\mathrm{NaBr + H_3PO_4 \xrightarrow{\Delta} NaH_2PO_4 + HBr\uparrow}$

$\mathrm{NaI + H_3PO_4 \xrightarrow{\Delta} NaH_2PO_4 + HI\uparrow}$

Alternatively, HBr and HI can be made by the hydrolysis of phosphorus tribromide or triiodide (red P + $\mathrm{Br_2}$ or $\mathrm{I_2}$ + water):

$\mathrm{PBr_3 + 3H_2O \rightarrow H_3PO_3 + 3HBr}$

Test for $\mathrm{Br^-}$ ions in aqueous solution

Acidify the solution with dilute $\mathrm{HNO_3}$ and add $\mathrm{AgNO_3}$ solution. A pale yellow precipitate of AgBr forms, which is sparingly soluble in dilute ammonia and fully soluble in conc. ammonia:

$\mathrm{Br^- + Ag^+ \rightarrow AgBr\downarrow\ (pale\ yellow)}$

Confirmation: add chlorine water and a little $\mathrm{CCl_4}$ and shake. The $\mathrm{CCl_4}$ layer turns orange-brown because $\mathrm{Br_2}$ is liberated: $\mathrm{2Br^- + Cl_2 \rightarrow 2Cl^- + Br_2}$.

Laboratory preparation of $\mathrm{SO_2}$

Sulphur dioxide is prepared in the laboratory by heating copper turnings with conc. $\mathrm{H_2SO_4}$:

$\mathrm{Cu + 2H_2SO_4(conc.) \xrightarrow{\Delta} CuSO_4 + SO_2\uparrow + 2H_2O}$

Alternatively, treat sodium sulphite (or bisulphite) with dilute $\mathrm{H_2SO_4}$ or $\mathrm{HCl}$:

$\mathrm{Na_2SO_3 + H_2SO_4 \rightarrow Na_2SO_4 + SO_2\uparrow + H_2O}$

The gas is dried by passing it through conc. $\mathrm{H_2SO_4}$ and collected by downward delivery (upward displacement of air), since it is about twice as dense as air and very soluble in water.

$\mathrm{SO_2}$ as a reducing agent (sulphur goes from $+4$ to $+6$):

$\mathrm{SO_2 + Cl_2 + 2H_2O \rightarrow H_2SO_4 + 2HCl}$

Also, acidified $\mathrm{KMnO_4}$ is decolourised:

$\mathrm{2KMnO_4 + 5SO_2 + 2H_2O \rightarrow K_2SO_4 + 2MnSO_4 + 2H_2SO_4}$

$\mathrm{SO_2}$ as an oxidising agent (sulphur goes from $+4$ to $0$):

$\mathrm{SO_2 + 2H_2S \rightarrow 3S\downarrow + 2H_2O}$

Here $\mathrm{H_2S}$ is oxidised to sulphur and $\mathrm{SO_2}$ is reduced, so $\mathrm{SO_2}$ shows both behaviours because sulphur is in an intermediate oxidation state ($+4$).

(a) Manufacture of nitric acid by Ostwald's process

Principle: Ammonia is catalytically oxidised by air to nitric oxide (NO) over a platinum-rhodium catalyst. NO is oxidised to $\mathrm{NO_2}$, which is absorbed in water in the presence of excess air to give $\mathrm{HNO_3}$.

Procedure:

1. Catalytic oxidation of ammonia. A dry, purified mixture of about $1$ part $\mathrm{NH_3}$ and $10$ parts air is passed through a converter containing a hot Pt-Rh gauze catalyst at about $800^\circ\mathrm{C}$ (the reaction is exothermic and keeps the catalyst hot):

$\mathrm{4NH_3 + 5O_2 \xrightarrow{Pt-Rh,\ 800^\circ C} 4NO + 6H_2O}$

2. Oxidation of NO. The hot gases are cooled and mixed with more air in an oxidation chamber:

$\mathrm{2NO + O_2 \rightarrow 2NO_2}$

3. Absorption. $\mathrm{NO_2}$ passes up an absorption tower containing quartz and is absorbed by water trickling down, with air supplied:

$\mathrm{4NO_2 + O_2 + 2H_2O \rightarrow 4HNO_3}$

The acid obtained is about $60$–$65\%$; it is concentrated by distillation with conc. $\mathrm{H_2SO_4}$ to about $68\%$ (constant-boiling mixture).

Diagram (block form): Dry air + $\mathrm{NH_3}$ $\rightarrow$ converter (Pt-Rh gauze, $800^\circ\mathrm{C}$) $\rightarrow$ cooler / oxidation chamber (+ air) $\rightarrow$ absorption tower (water in at the top, $\mathrm{HNO_3}$ out at the bottom).

Conc. $\mathrm{HNO_3}$ with:

(i) KI - the iodide is oxidised to iodine and brown fumes of $\mathrm{NO_2}$ are evolved:

$\mathrm{2KI + 4HNO_3(conc.) \rightarrow 2KNO_3 + I_2 + 2NO_2\uparrow + 2H_2O}$

(ii) Zn - with conc. acid, $\mathrm{NO_2}$ is produced:

$\mathrm{Zn + 4HNO_3(conc.) \rightarrow Zn(NO_3)_2 + 2NO_2\uparrow + 2H_2O}$

(b) Explain why

(i) Conc. $\mathrm{HNO_3}$ decomposes slowly in sunlight to give $\mathrm{NO_2}$, which makes the acid yellow-brown:

$\mathrm{4HNO_3 \xrightarrow{light} 4NO_2 + 2H_2O + O_2}$

It is therefore stored in a dark (amber) bottle to prevent this decomposition.

(ii) Conc. $\mathrm{HNO_3}$ oxidises the surface of iron to form a thin, tough, unreactive layer of oxide ($\mathrm{Fe_3O_4}$ / $\mathrm{Fe_2O_3}$) that is not attacked further and protects the metal below. The iron is then said to be passive, and it no longer reacts with the acid (or with $\mathrm{CuSO_4}$ solution).

Because concentrated sulphuric acid is a strong oxidising agent, it oxidises hydrogen iodide to iodine; the reaction does not give hydrogen iodide.

$\mathrm{2NaI + 2H_{2}SO_{4} \rightarrow Na_{2}SO_{4} + SO_{2} + I_{2} + 2H_{2}O}$

So the gas obtained is not HI but a mixture of $\mathrm{SO_{2}}$ and $\mathrm{I_{2}}$ vapour. (For HCl, sulphuric acid is only a non-volatile acid, so no oxidation occurs.)

Hydrogen chloride gas vs hydrochloric acid

So the essential difference is that HCl gas is a dry covalent compound with no acidic behaviour, while hydrochloric acid is its aqueous solution which ionises to give H⁺ ions and therefore behaves as an acid.

Identification of HCl bottle using ammonia gas:

Bring a glass rod dipped in concentrated ammonia solution near the mouth of each bottle. The bottle containing HCl will give dense white fumes of ammonium chloride.

$\mathrm{NH_{3} + HCl \rightarrow NH_{4}Cl}$

The other bottles give no such white fumes, so the bottle containing HCl is identified.

i. Classification of the given oxides:

So, from the table:

• Acidic oxide: SO$_2$
• Basic oxide: Na$_2$O
• Neutral oxide: CO
• Mixed oxide: Fe$_3$O$_4$

(Al$_2$O$_3$ is amphoteric and H$_2$O$_2$ is a peroxide.)

Amphoteric oxide: aluminium oxide (Al₂O₃)

An amphoteric oxide reacts with both acids and bases to form salt and water. Al₂O₃ behaves as a base with acid and as an acid with base:

1. With acid (behaves as a base):

$\mathrm{Al_{2}O_{3} + 6HCl \rightarrow 2AlCl_{3} + 3H_{2}O}$

2. With base (behaves as an acid):

$\mathrm{Al_{2}O_{3} + 2NaOH \rightarrow 2NaAlO_{2} + H_{2}O}$

Since Al₂O₃ reacts with both HCl and NaOH, it is amphoteric in nature.

It binds strongly to haemoglobin, forming carboxyhaemoglobin, and prevents haemoglobin from carrying oxygen to body tissues.

Oxygen is used in the manufacture of steel (in the Bessemer process and basic oxygen furnace), where it is blown through molten pig iron to burn off carbon and other impurities.

Sulphuric acid as a precipitant (it precipitates a salt from solution):

$\mathrm{BaCl_{2} + H_{2}SO_{4} \rightarrow BaSO_{4} \downarrow + 2HCl}$

Here dilute H₂SO₄ precipitates white insoluble barium sulphate.

Sulphuric acid as a dehydrating agent (it removes water from a compound):

$\mathrm{C_{6}H_{12}O_{6} \rightarrow [conc. H_{2}SO_{4}] 6C + 6H_{2}O}$

Here concentrated H₂SO₄ dehydrates glucose to give black carbon.

Phosphatic fertilizer (single superphosphate) is made by treating calcium phosphate (phosphate rock) with sulphuric acid:

$\text{Ca}_3(\text{PO}_4)_2 + 2\text{H}_2\text{SO}_4 \rightarrow \text{Ca}(\text{H}_2\text{PO}_4)_2 + 2\text{CaSO}_4$

Similarly, ammonium sulphate fertilizer is obtained from ammonia and sulphuric acid:

$2\text{NH}_3 + \text{H}_2\text{SO}_4 \rightarrow (\text{NH}_4)_2\text{SO}_4$

Because the sulphur in H₂SO₄ is in its highest oxidation state (+6), so it can only gain electrons (be reduced) and therefore acts as an oxidizing agent.

In H₂SO₄, sulphur is at its maximum oxidation state of +6. An element in its highest oxidation state cannot be oxidized further — it can only accept electrons. Hence H₂SO₄ can only undergo reduction, which means it always behaves as an oxidizing agent (especially hot concentrated H₂SO₄, which oxidizes Cu, C, S, etc.).