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Chemistry - Study of Compounds (HCl, Ammonia, Nitric Acid, Sulphuric Acid)

Grade 10ICSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Hydrogen Chloride (HCl): Lab preparation involves the reaction of sodium chloride with concentrated sulphuric acid: NaCl+H2SO4→<200∘CNaHSO4+HClNaCl + H_2SO_4 \xrightarrow{< 200^\circ C} NaHSO_4 + HCl. It is a polar covalent compound and is highly soluble in water, demonstrated by the Fountain Experiment.

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Ammonia (NH3): Prepared in the lab by heating a metal hydroxide with an ammonium salt, e.g., Ca(OH)2+2NH4Cl→CaCl2+2H2O+2NH3Ca(OH)_2 + 2NH_4Cl \rightarrow CaCl_2 + 2H_2O + 2NH_3. Industrially, it is produced via the Haber's Process using N2N_2 and H2H_2.

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Nitric Acid (HNO3): Prepared in a glass retort using concentrated H2SO4H_2SO_4 and KNO3KNO_3 or NaNO3NaNO_3. Industrially produced via the Ostwald's Process. It is a powerful oxidizing agent and exhibits the 'Brown Ring Test' for nitrate ions.

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Sulphuric Acid (H2SO4): Known as the 'King of Chemicals'. Produced by the Contact Process. It acts as a strong dibasic acid, a non-volatile acid, a dehydrating agent (removes chemically combined water), and an oxidizing agent.

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Physical Properties: HClHCl and NH3NH_3 are both highly soluble in water. NH3NH_3 is the only common alkaline gas. HNO3HNO_3 usually appears yellowish due to the presence of dissolved NO2NO_2 gas.

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Chemical Tests: HClHCl gas gives dense white fumes of NH4ClNH_4Cl when a rod dipped in NH3NH_3 solution is brought near it. Concentrated H2SO4H_2SO_4 chars sugar to a black spongy mass of carbon.

📐Formulae

N2(g)+3H2(g)⇌Fe+Mo450−500∘C,200atm2NH3(g)+HeatN_2(g) + 3H_2(g) \xrightleftharpoons[Fe+Mo]{450-500^\circ C, 200 atm} 2NH_3(g) + \text{Heat}

4NH3+5O2→Pt,800∘C4NO+6H2O+Heat4NH_3 + 5O_2 \xrightarrow{Pt, 800^\circ C} 4NO + 6H_2O + \text{Heat}

2SO2+O2⇌V2O5450∘C2SO32SO_2 + O_2 \xrightleftharpoons[V_2O_5]{450^\circ C} 2SO_3

H2SO4+SO3→H2S2O7 (Oleum)H_2SO_4 + SO_3 \rightarrow H_2S_2O_7 \text{ (Oleum)}

Cu+4HNO3 (conc.)→Cu(NO3)2+2NO2+2H2OCu + 4HNO_3 \text{ (conc.)} \rightarrow Cu(NO_3)_2 + 2NO_2 + 2H_2O

C12H22O11→conc.H2SO412C+11H2OC_{12}H_{22}O_{11} \xrightarrow{conc. H_2SO_4} 12C + 11H_2O

💡Examples

Problem 1:

Explain why concentrated H2SO4H_2SO_4 is used in the preparation of HClHCl and HNO3HNO_3 but not NH3NH_3.

Solution:

Concentrated H2SO4H_2SO_4 is used for HClHCl and HNO3HNO_3 because it is a non-volatile acid and can displace more volatile acids from their salts. It is not used for NH3NH_3 because NH3NH_3 is basic and would react with the acid to form a salt.

Explanation:

Reaction for NH3NH_3: 2NH3+H2SO4→(NH4)2SO42NH_3 + H_2SO_4 \rightarrow (NH_4)_2SO_4. Hence, CaOCaO (Quicklime) is used as a drying agent for NH3NH_3 instead.

Problem 2:

State the observation when NH3NH_3 gas is passed over heated CuOCuO.

Solution:

The black copper(II) oxide (CuOCuO) is reduced to a reddish-brown solid (metallic copper).

Explanation:

The chemical equation is: 3CuO+2NH3→3Cu+3H2O+N23CuO + 2NH_3 \rightarrow 3Cu + 3H_2O + N_2. NH3NH_3 acts as a reducing agent in this reaction.

Problem 3:

What is the 'Brown Ring Test' and what does it detect?

Solution:

It detects the nitrate ion (NO3−NO_3^-). A freshly prepared FeSO4FeSO_4 solution is added to a substance, and concentrated H2SO4H_2SO_4 is poured along the sides. A brown ring forms at the junction.

Explanation:

The ring is due to the formation of nitroso-ferrous sulphate: [Fe(H2O)5(NO)]SO4[Fe(H_2O)_5(NO)]SO_4.