Chemical Reactions Class 10 Previous Year Questions

Step 1: Write the reaction: \( CaO(s) + H_2O(l) \rightarrow Ca(OH)_2(aq) \).

Step 2: This reaction releases a large amount of heat, so it is exothermic.

Step 3: The product, calcium hydroxide, is a base dissolved in water, so its pH is more than 7.

Final answer: Option (D) — exothermic and the pH of the solution formed is more than 7.

Step 1 (temperature change): Calcium oxide reacting with water releases heat, so the mixture becomes warm.

\[ CaO(s) + H_2O(l) \rightarrow Ca(OH)_2(aq) + \text{Heat} \]

Step 2 (colour change): Green ferrous sulphate crystals turn brown on heating as they decompose.

\[ 2FeSO_4(s) \xrightarrow{\text{Heat}} Fe_2O_3(s) + SO_2(g) + SO_3(g) \]

Step 3 (precipitate formation): Barium chloride solution mixed with sodium sulphate solution gives an insoluble white solid.

\[ BaCl_2(aq) + Na_2SO_4(aq) \rightarrow BaSO_4(s) + 2NaCl(aq) \]

Final answer: Each equation above is labelled with the exact change it demonstrates — this labelling is what a marking scheme checks for, not just the equation.

Step 1: In options (a), (b) and (c), the identity of the substance changes — souring, fermentation and rusting all form new substances.

Step 2: Melting of glaciers is only a change of state, from solid ice to liquid water. The substance is still \( H_2O \); nothing new is formed.

Final answer: Option (d) — melting of glaciers is a physical change, not a chemical reaction.

Step 1: On heating, the surface of copper powder becomes coated black. This is caused by oxidation — copper gains oxygen from the air.

\[ 2Cu(s) + O_2(g) \xrightarrow{\text{Heat}} 2CuO(s) \]

Step 2: This is a fast reaction driven by heat, producing black copper(II) oxide directly.

Step 3: Copper left in open air is not heated. It reacts slowly with moisture and gases present in air over time, and a green coating forms on its surface instead of the black oxide.

Final answer: Heating gives fast oxidation to black \( CuO \); ordinary corrosion in open air is a slow reaction at room temperature that leaves a green coating — the two are different rates and different products of attack by oxygen and moisture.

Step 1: Silver chloride decomposes only when light energy is supplied — this is the essential condition.

\[ 2AgCl(s) \xrightarrow{\text{Sunlight}} 2Ag(s) + Cl_2(g) \]

Step 2: This is a decomposition reaction — one reactant breaks into simpler products under light, so it is called a photochemical decomposition.

Step 3: Ferrous sulphate crystals decompose only when heated — the essential condition is heat.

\[ 2FeSO_4(s) \xrightarrow{\Delta} Fe_2O_3(s) + SO_2(g) + SO_3(g) \]

Final answer: Both are decomposition reactions; AgCl needs sunlight (photochemical decomposition) and FeSO4 needs heat (thermal decomposition).

Step 1 (colour change): Ferrous sulphate crystals change from green towards brown on heating as they decompose.

\[ 2FeSO_4(s) \xrightarrow{\Delta} Fe_2O_3(s) + SO_2(g) + SO_3(g) \]

Step 2 (temperature change/rise): Calcium oxide reacting with water gives out heat.

\[ CaO(s) + H_2O(l) \rightarrow Ca(OH)_2(aq) + \text{Heat} \]

Step 3 (precipitate): Barium chloride and sodium sulphate solutions give an insoluble white solid.

\[ BaCl_2(aq) + Na_2SO_4(aq) \rightarrow BaSO_4(s) + 2NaCl(aq) \]

Final answer: The three equations above, each with its named change, complete the answer — the change and the equation must be stated side by side.

Step 1: A reaction in which two compounds exchange ions between the reactants to form two new compounds is a double displacement reaction, as seen when \( Na_2SO_4 \) and \( BaCl_2 \) exchange their sulphate and chloride ions.

Final answer: Option (C) — Double Displacement Reaction.

Step 1: On heating, limestone (calcium carbonate) breaks down into calcium oxide and carbon dioxide.

\[ CaCO_3(s) \xrightarrow{\text{Heat}} CaO(s) + CO_2(g) \]

Step 2: Calcium oxide is commonly called quick lime.

Final answer: Option (A) — Quick lime.

Step 1 (gas evolution): Zinc granules added to dilute sulphuric acid release hydrogen gas.

\[ Zn(s) + H_2SO_4(aq) \rightarrow ZnSO_4(aq) + H_2(g) \]

Step 2 (colour change): An iron nail dipped in copper sulphate solution turns brownish, and the blue colour of the solution fades.

\[ Fe(s) + CuSO_4(aq) \rightarrow FeSO_4(aq) + Cu(s) \]

Final answer: Both examples above are correct answers for gas evolution and colour change respectively.

Step 1 (Option a): Copper powder heated in a china dish turns black as copper(II) oxide forms.

\[ 2Cu(s) + O_2(g) \xrightarrow{\text{Heat}} 2CuO(s) \]

Step 2 (Option b): Barium chloride solution reacts with sodium sulphate solution to give a white precipitate of barium sulphate.

\[ BaCl_2(aq) + Na_2SO_4(aq) \rightarrow BaSO_4(s) + 2NaCl(aq) \]

Step 3: The ions present in the precipitated compound are \( Ba^{2+} \) and \( SO_4^{2-} \); sodium and chloride ions remain dissolved as \( Na^+ \) and \( Cl^- \).

Final answer: Either route earns full marks: black copper(II) oxide by heating, or barium sulphate precipitate with \( Ba^{2+} \) and \( SO_4^{2-} \) identified.

Step 1: Magnesium ribbon burns in air with a dazzling white flame, forming white magnesium oxide ash.

\[ 2Mg(s) + O_2(g) \rightarrow 2MgO(s) \]

Step 2: Two reactants, magnesium and oxygen, combine to form a single product, magnesium oxide. This makes it a combination reaction.

Step 3: Heat and light are given out during the reaction, which also makes it exothermic.

Final answer: It is a combination reaction, because two reactants join to give a single product, magnesium oxide.

Step 1: Write the equation: \( CaO(s) + H_2O(l) \rightarrow Ca(OH)_2(aq) \).

Step 2: Two reactants, calcium oxide and water, combine to give a single product, calcium hydroxide.

Final answer: It is a combination reaction; naming the type without this equation and reasoning would earn only partial credit under the marking scheme.

Option (a) — activities for colour change and temperature change: Heating ferrous sulphate crystals shows a colour change from green towards brown as \( 2FeSO_4(s) \xrightarrow{\Delta} Fe_2O_3(s) + SO_2(g) + SO_3(g) \). Adding water to calcium oxide shows a temperature rise, felt on touching the container, as \( CaO(s) + H_2O(l) \rightarrow Ca(OH)_2(aq) \).

Option (b), part (i) — decomposition reaction: A decomposition reaction is one where a single reactant breaks down into two or more simpler products. Electrolysis of water uses electrical energy to split water into hydrogen and oxygen gas. Silver bromide decomposes into silver and bromine when exposed to sunlight, using light energy.

Option (b), part (ii) — opposite reactions: Calcium oxide is formed when calcium carbonate decomposes on heating: \( CaCO_3(s) \xrightarrow{\text{Heat}} CaO(s) + CO_2(g) \). This single substance splitting into two products is a decomposition reaction. Calcium hydroxide is formed when calcium oxide combines with water: \( CaO(s) + H_2O(l) \rightarrow Ca(OH)_2(aq) \). This is two substances joining into one, a combination reaction — the exact reverse process. That is why the two reactions are called opposite to each other.

Final answer: Both option (a) and option (b) are complete answers on their own; a student only needs to attempt one in full.

Step 1: A building material that is insoluble in water and reacts with acid to release a gas is calcium carbonate, commonly found as marble or limestone. So \( X = CaCO_3 \).

Step 2: Write the reaction of X with dilute hydrochloric acid.

\[ CaCO_3(s) + 2HCl(aq) \rightarrow CaCl_2(aq) + H_2O(l) + CO_2(g) \]

Step 3: The gas released, carbon dioxide, turns lime water milky because it forms an insoluble precipitate.

\[ Ca(OH)_2(aq) + CO_2(g) \rightarrow CaCO_3(s) + H_2O(l) \]

Final answer: X is calcium carbonate \( (CaCO_3) \); the two equations above show its reaction with acid and the lime water test on the gas produced.

Step 1: Before exposure to sunlight, silver chloride is white.

Step 2: On exposure to sunlight, it decomposes into grey silver metal and chlorine gas — a photochemical decomposition reaction.

\[ 2AgCl(s) \xrightarrow{\text{Sunlight}} 2Ag(s) + Cl_2(g) \]

Step 3: This light sensitivity is used in black and white photography. Silver bromide \( (AgBr) \) behaves the same way and can be used for the same purpose.

Final answer: Silver chloride was white; it decomposes in sunlight (photochemical decomposition) to silver and chlorine; it is used in black and white photography, and silver bromide is the alternative chemical for the same use.

Step 1: Check what each reactant loses or gains. \( MnO_2 \) loses oxygen to become \( MnCl_2 \), so manganese dioxide is reduced.

Step 2: \( HCl \) loses hydrogen and forms \( Cl_2 \), so hydrochloric acid is oxidised.

Step 3: Since one reactant is oxidised while the other is reduced in the same reaction, it fits the definition of a redox reaction.

Final answer: Yes, it is a redox reaction — \( HCl \) is oxidised to \( Cl_2 \) and \( MnO_2 \) is reduced to \( MnCl_2 \).

Step 1: Write the word equation: sodium carbonate + calcium chloride → calcium carbonate + sodium chloride.

Step 2: Write the skeletal equation using formulae: \( Na_2CO_3 + CaCl_2 \rightarrow CaCO_3 + NaCl \).

Step 3: Count atoms on each side. Na: 2 on the left, 1 on the right. Cl: 2 on the left, 1 on the right. Ca, C and O already match.

Step 4: Fix Na and Cl together by placing a coefficient of 2 before \( NaCl \).

\[ Na_2CO_3 + CaCl_2 \rightarrow CaCO_3 + 2NaCl \]

Step 5: Recheck every element: Na (2=2), C (1=1), O (3=3), Ca (1=1), Cl (2=2). The equation is balanced.

Step 6: Add state symbols, since both starting solutions are in water and calcium carbonate is the insoluble product.

\[ Na_2CO_3(aq) + CaCl_2(aq) \rightarrow CaCO_3(s) + 2NaCl(aq) \]

Final answer: This is a double displacement (precipitation) reaction — the \( Ca^{2+} \) and \( CO_3^{2-} \) ions exchange partners with \( Na^+ \) and \( Cl^- \), and the insoluble calcium carbonate settles out as a precipitate.