CBSE Class 10 Science Advanced Chemistry Chemical Reactions and Equations Short Answer Questions Set 01

Short Answer Type

Question 1. What do you mean by a precipitation reaction ? Explain by giving examples.
Answer:
A reaction in which an insoluble solid (precipitate) is formed on mixing two solutions is called a precipitation reaction.
Example 1: \( \ce{Na2CO3(aq) + CaCl2(aq) -> CaCO3(s) + 2NaCl(aq)} \). Calcium carbonate is the precipitate.
Example 2: \( \ce{Na2SO4(aq) + BaCl2(aq) -> BaSO4(s) + 2NaCl(aq)} \). Barium sulphate is the white precipitate.
In simple words: When two clear solutions are mixed and a solid appears, it is a precipitation reaction.

Teacher's Note:
1. The precipitate is shown by (s) or a downward arrow.
2. Such reactions are also double displacement reactions.

Question 2. Write the balanced equation for the following chemical reactions. (i) Hydrogen + Chlorine → Hydrogen chloride (ii) Barium chloride + Aluminium sulphate → Barium sulphate + Aluminium chloride (iii) Sodium + Water → Sodium hydroxide + Hydrogen
Answer:
(i) \( \ce{H2(g) + Cl2(g) -> 2HCl(g)} \)
(ii) \( \ce{3BaCl2 + Al2(SO4)3 -> 3BaSO4 + 2AlCl3} \) (in solution, with \( \ce{BaSO4} \) as a precipitate)
(iii) \( \ce{2Na(s) + 2H2O(l) -> 2NaOH(aq) + H2(g)} \)
In simple words: Write the formulae first and then put coefficients until each element is equal on both sides.

Teacher's Note:
1. Never change a subscript to balance an equation; change only coefficients.
2. Balance the element that appears in the fewest formulae first.

Question 3. Write a balanced chemical equation with state symbols for the following reactions. (i) Solutions of barium chloride and sodium sulphate in water react to give insoluble barium sulphate and the solution of sodium chloride. (ii) Sodium hydroxide solution (in water) reacts with hydrochloric acid solution (in water) to produce sodium chloride solution and water.
Answer:
(i) \( \ce{BaCl2(aq) + Na2SO4(aq) -> BaSO4(s) + 2NaCl(aq)} \)
(ii) \( \ce{NaOH(aq) + HCl(aq) -> NaCl(aq) + H2O(l)} \)
(The book writes sodium chloride in (ii) as solid, but the question calls it a solution, so (aq) is used.)
In simple words: A substance dissolved in water is (aq), an insoluble solid is (s) and water itself is (l).

Teacher's Note:
1. Read the words 'insoluble' and 'solution' in the question to choose the state symbols.
2. Both of these are double displacement reactions.

Question 4. Action of water on calcium carbide (\( \ce{CaC2} \)) is used for the preparation of acetylene gas (\( \ce{C2H2} \)) used in welding. After the reaction, calcium carbide is converted into slaked lime. Write the balanced equation for the reaction.
Answer:
\( \ce{CaC2 + 2H2O -> Ca(OH)2 + C2H2 ^} \)
Check: Ca 1 and 1, C 2 and 2, H 4 and 4, O 2 and 2.
In simple words: Calcium carbide plus two molecules of water gives slaked lime and acetylene gas.

Teacher's Note:
1. Slaked lime is calcium hydroxide.
2. Always check by counting each element on both sides.

Question 5. A pinch of solid sodium carbonate is added into dilute hydrochloric acid in a test tube. Sodium chloride and water formed remain in the solution whereas carbon dioxide escapes out. The tube is found to become hot. How will the reaction be represented to make it maximum informative ?
Answer:
\( \ce{Na2CO3(s) + 2HCl "(dil)" -> 2NaCl(aq) + H2O(l) + CO2(g) ^ + "heat"} \)
This equation shows the physical states, the dilute acid, the escaping gas, and that heat is released (exothermic).
In simple words: A good equation tells the states, the concentration, the gas given off and the heat released.

Teacher's Note:
1. Add (s), (l), (g), (aq), dilute or conc., arrows for gas and precipitate, and heat as needed.
2. Balance first, then add the extra information.

Question 6. Balance the following chemical equations. (a) \( \ce{Al + O2 -> Al2O3} \) (b) \( \ce{C4H10 + O2 -> CO2 + H2O} \)
Answer:
(a) \( \ce{4Al + 3O2 -> 2Al2O3} \)
(b) \( \ce{2C4H10 + 13O2 -> 8CO2 + 10H2O} \)
Check (b): C 8 and 8, H 20 and 20, O 26 and 16 + 10 = 26.
In simple words: Fix the carbon and hydrogen first and keep oxygen for the last.

Teacher's Note:
1. When the oxygen count is odd, double all other coefficients.
2. Write each element count to avoid slips.

Question 7. Write the balanced chemical equation for the following and identify the type of reaction in each case. (a) Potassium bromide(aq) + Barium iodide(aq) → Potassium iodide(aq) + Barium bromide(s) (b) Zinc carbonate(s) → Zinc oxide(s) + Carbon dioxide(g) (c) Hydrogen(g) + Chlorine(g) → Hydrogen chloride(g) (d) Magnesium(s) + Hydrochloric acid(aq) → Magnesium chloride(aq) + Hydrogen(g)
Answer:
(a) \( \ce{2KBr(aq) + BaI2(aq) -> 2KI(aq) + BaBr2(s)} \): double displacement reaction.
(b) \( \ce{ZnCO3(s) -> ZnO(s) + CO2(g)} \): decomposition reaction.
(c) \( \ce{H2(g) + Cl2(g) -> 2HCl(g)} \): combination reaction.
(d) \( \ce{Mg(s) + 2HCl(aq) -> MgCl2(aq) + H2(g)} \): displacement reaction.
In simple words: Two compounds swapping ions is double displacement, one compound splitting is decomposition, two substances joining is combination, and an element replacing another is displacement.

Teacher's Note:
1. In real life barium bromide is soluble, but the question gives it as (s) and we follow it.
2. Check the number of reactants and products to find the type.

Question 8. Identify the substances that are oxidised and the substances that are reduced in the following reactions. (i) \( \ce{4Na(s) + O2(g) -> 2Na2O(s)} \) (ii) \( \ce{CuO(s) + H2(g) -> Cu(s) + H2O(l)} \)
Answer:
(i) Sodium gains oxygen, so Na is oxidised, and oxygen is reduced.
(ii) CuO loses oxygen, so CuO is reduced to Cu, and hydrogen gains oxygen, so H₂ is oxidised to H₂O.
In simple words: The one that gains oxygen is oxidised, and the one that loses oxygen is reduced.

Teacher's Note:
1. Oxidation and reduction go together in a redox reaction.
2. Also confirm with electrons: Na loses electrons and O gains them.

Question 9. Calculate the oxidation number of (i) S in \( \ce{H2SO4} \) (ii) N in \( \ce{(NH4)2SO4} \).
Answer:
(i) Let the oxidation number of S be x: 2(+1) + x + 4(-2) = 0, so x = +6.
(ii) In the ammonium ion \( \ce{NH4+} \): x + 4(+1) = +1, so x = -3. (Check for the whole compound: 2(-3) + 8(+1) + (+6) + 4(-2) = 0.)
In simple words: Add the known oxidation numbers and solve for x so that the total equals the charge on the particle.

Teacher's Note:
1. Use +1 for H and -2 for O in these compounds.
2. For an ion, the sum equals the charge, and for a neutral compound it equals zero.

Question 10. On heating blue coloured powder of copper(II) nitrate in a boiling tube, copper oxide (black), oxygen gas and a brown gas X is formed. (a) Write a balanced chemical equation of the reaction. (b) Identify the brown gas X evolved. (c) Identify the type of reaction. (d) What could be the nature of aqueous solution of the gas X ?
Answer:
(a) \( \ce{2Cu(NO3)2 ->[\Delta] 2CuO + 4NO2 + O2} \)
(b) X is nitrogen dioxide (\( \ce{NO2} \)).
(c) It is a thermal decomposition reaction.
(d) The solution of \( \ce{NO2} \) in water is acidic, because it forms nitric and nitrous acids: \( \ce{2NO2 + H2O -> HNO3 + HNO2} \).
In simple words: Blue copper nitrate decomposes on heating into black CuO, brown fumes of NO2 and oxygen.

Teacher's Note:
1. Brown fumes are the tell-tale sign of NO2.
2. Acidic oxides of non-metals give acidic solutions.

Question 11. In the following reaction, name the substance oxidized and the substance reduced : \( \ce{PbS(s) + 4H2O2(aq) -> PbSO4(s) + 4H2O(l)} \) Which substance acts as oxidizing agent and which one as reducing agent ?
Answer:
\( \ce{PbS} \) gains oxygen (S changes from -2 to +6), so PbS is oxidised and acts as the reducing agent.
\( \ce{H2O2} \) loses oxygen to become water, so H₂O₂ is reduced and is the oxidising agent.
In simple words: PbS takes oxygen from hydrogen peroxide, so PbS is oxidised and hydrogen peroxide is reduced.

Teacher's Note:
1. The substance oxidised is the reducing agent, and the substance reduced is the oxidising agent.
2. This reaction is used to restore old blackened lead paintings.

Question 12. Why does a zinc plate develop holes when placed in copper sulphate solution for a few hours/days ?
Answer:
Zinc is more reactive than copper, so it displaces copper from copper sulphate: \( \ce{Zn(s) + CuSO4(aq) -> ZnSO4(aq) + Cu(s)} \).
Zinc keeps dissolving as zinc sulphate, so the plate is slowly eaten away and develops holes, while copper deposits as a brown layer.
In simple words: Zinc goes into the solution, leaving holes in the plate.

Teacher's Note:
1. The blue colour of the solution slowly fades.
2. More reactive metals displace the less reactive ones from their salt solutions.

Question 13. (i) What is observed when a solution of potassium iodide is added to a solution of lead nitrate taken in a test tube ? (ii) What type of reaction is this ? (iii) Write a balanced chemical equation to represent the above reaction.
Answer:
(i) A yellow precipitate of lead iodide is formed.
(ii) It is a double displacement (precipitation) reaction.
(iii) \( \ce{Pb(NO3)2(aq) + 2KI(aq) -> PbI2(s) + 2KNO3(aq)} \)
In simple words: Two colourless solutions give a bright yellow solid.

Teacher's Note:
1. Yellow PbI2 is a famous colour to remember for exams.
2. Do not mix it up with white PbCl2.

Question 14. (a) What is the name and formula of the product formed when water is added to quick lime for white wash ? (b) After white washing the walls, why do the walls begin to shine after 2-3 days ?
Answer:
(a) Calcium hydroxide (slaked lime), \( \ce{Ca(OH)2} \): \( \ce{CaO + H2O -> Ca(OH)2} \).
(b) Calcium hydroxide slowly reacts with carbon dioxide of the air and forms a thin, shiny layer of calcium carbonate: \( \ce{Ca(OH)2 + CO2 -> CaCO3 + H2O} \). This gives the walls a shine.
In simple words: Slaked lime turns into shiny calcium carbonate by reacting with carbon dioxide of the air.

Teacher's Note:
1. The finish takes two to three days, as the reaction is slow.
2. Marble is also calcium carbonate, which is why the walls look glossy.

Question 15. A silver article generally turns black when kept in the open for a few days. The article when rubbed with toothpaste again starts shining. (a) Why do silver articles turn black when kept in the open for a few days ? Name the phenomenon involved. (b) Name the black substance formed and give its chemical formula.
Answer:
(a) Silver reacts with hydrogen sulphide present in the air and gets a black coating. This slow attack of the air on a metal is corrosion (tarnishing).
(b) The black substance is silver sulphide, \( \ce{Ag2S} \): \( \ce{2Ag + H2S -> Ag2S + H2} \).
In simple words: Silver slowly reacts with gases in the air and turns black; toothpaste rubs the layer away.

Teacher's Note:
1. Copper gets a green coating and iron gets rust in the same way.
2. Toothpaste works as a mild abrasive.

Teacher's Note:
1. In (e) one of the lead atoms goes from +4 to +2 and chloride goes to chlorine.
2. Use the change of oxidation numbers when oxygen or hydrogen is not obvious.

Question 17. During the reaction of some metals with dilute hydrochloric acid, the following observations were made. (a) Silver metal does not show any change (b) The reaction of sodium metal is found to be highly explosive (c) Some bubbles of a gas are seen when lead (Pb) is reacted with the acid. Explain these observations by giving suitable reasons.
Answer:
The behaviour depends on the position of the metal in the reactivity series: Na > Zn > Pb > H > Cu > Ag.
(a) Silver is below hydrogen, so it cannot displace hydrogen from the acid and nothing happens.
(b) Sodium is highly reactive; it displaces hydrogen very quickly and the heat released ignites the hydrogen, so the reaction is explosive.
(c) Lead is only slightly above hydrogen, so it reacts slowly and gives a few bubbles of hydrogen gas.
In simple words: The more reactive the metal, the faster it pushes hydrogen out of the acid.

Teacher's Note:
1. Only metals above hydrogen in the reactivity series give hydrogen with dilute acids.
2. Because of the insoluble coating of PbCl2, lead reacts very slowly.

Question 18. Write a balanced chemical equation for each of the following reactions and also classify them. (a) Lead acetate solution is treated with dilute hydrochloric acid to form lead chloride and acetic acid solution. (b) Iron(III) oxide on heating with carbon monoxide gas reacts to form solid iron and liberates carbon dioxide gas.
Answer:
(a) \( \ce{Pb(CH3COO)2(aq) + 2HCl(aq) -> PbCl2(s) + 2CH3COOH(aq)} \): a double displacement (precipitation) reaction.
(b) \( \ce{Fe2O3(s) + 3CO(g) -> 2Fe(s) + 3CO2(g)} \): a redox reaction (Fe₂O₃ is reduced and CO is oxidised).
In simple words: In (a) the ions exchange, and in (b) oxygen is passed from iron oxide to carbon monoxide.

Teacher's Note:
1. Check that atoms are equal on both sides, especially oxygen in (b).
2. A reaction can fall into more than one class.

Question 19. (a) What happens when copper is burned in air ? Give the equation. What type of a reaction is it ? (b) What happens when hydrogen gas is passed over the product obtained in step above ? Give the equation also. [CBSE 10th Board 2014, 3 Marks]
Answer:
(a) Shiny brown copper turns black because copper(II) oxide forms: \( \ce{2Cu + O2 ->[\Delta] 2CuO} \). It is a combination (oxidation) reaction.
(b) The black copper oxide is reduced and turns brown again, and water is formed: \( \ce{CuO + H2 ->[\Delta] Cu + H2O} \).
In simple words: Copper turns black by combining with oxygen, and hydrogen changes the black powder back to brown copper.

Teacher's Note:
1. The first reaction is oxidation and the second is reduction.
2. Mention colour changes in the answer, as they earn marks.

Question 20. Complete the following reactions and write the balanced equations. FeSO₄(s) on heating (decomposition) and Pb(NO₃)₂(s) on heating (decomposition). [CBSE 10th Board 2014, 3 Marks]
Answer:
\( \ce{2FeSO4(s) ->[\Delta] Fe2O3(s) + SO2(g) + SO3(g)} \) (green crystals leave a reddish brown residue and the smell of burning sulphur is given off).
\( \ce{2Pb(NO3)2(s) ->[\Delta] 2PbO(s) + 4NO2(g) + O2(g)} \) (white solid gives yellow residue and brown fumes).
In simple words: Heating each salt breaks it into an oxide and gases.

Teacher's Note:
1. Remember the colour changes of both reactions.
2. Both are thermal decomposition reactions.

Question 21. A solution of potassium chloride when mixed with silver nitrate solution, an insoluble white substance is formed. Write the chemical reaction involved and also mention the type of the chemical reaction. [CBSE 10th Board 2015, 2 Marks]
Answer:
\( \ce{KCl(aq) + AgNO3(aq) -> AgCl(s) + KNO3(aq)} \)
The white insoluble substance is silver chloride. It is a double displacement (precipitation) reaction.
In simple words: Silver chloride, a white solid, is formed by an exchange of ions.

Teacher's Note:
1. The reaction is just like the NaCl and AgNO3 reaction.
2. Name the white substance in your answer.

Question 22. "Respiration is an exothermic reaction." Justify this statement giving the chemical equation for the reaction involved. [CBSE 10th Board 2017, 2 Marks]
Answer:
In the cells, glucose is oxidised by the oxygen we breathe in, and energy is released: \( \ce{C6H12O6(aq) + 6O2(g) -> 6CO2(g) + 6H2O(l) + "energy"} \).
Since energy (heat) is released, respiration is an exothermic reaction. The heat keeps the body warm and works the muscles.
In simple words: Our cells burn glucose slowly with oxygen and give out energy.

Teacher's Note:
1. Respiration is a slow oxidation of food in cells, unlike burning, there is no flame.
2. Write the full equation with energy on the right side.

Question 23. (a) Identify the substance oxidized, substance reduced, oxidizing agent and reducing agent in the following reaction : \( \ce{ZnO + C -> Zn + CO} \) (b) Packets of potato chips are flushed with nitrogen gas, why ? [CBSE 10th Board 2017, 3 Marks]
Answer:
(a) Carbon gains oxygen, so C is oxidised and it is the reducing agent. ZnO loses oxygen, so ZnO is reduced and it is the oxidising agent.
(b) Nitrogen is an unreactive gas. It displaces air (oxygen) from the packet, so the oil in the chips is not oxidised and the chips do not turn rancid.
In simple words: Carbon takes oxygen from zinc oxide, and nitrogen in the packet keeps oxygen away from the chips.

Teacher's Note:
1. Part (a) is a perfect example of a redox reaction.
2. Say 'rancidity' in part (b) because it is the key word.

Question 24. When a white compound 'X' is placed under sunlight, it turns grey. Give the name of reaction and write the balanced chemical equation.
Answer:
The compound X is silver chloride (AgCl). The reaction is a photodecomposition (photolytic decomposition) reaction, in which light breaks the compound.
\( \ce{2AgCl(s) ->[sunlight] 2Ag(s) + Cl2(g)} \)
The grey colour is due to silver metal.
In simple words: Sunlight breaks white AgCl into grey silver and chlorine gas.

Teacher's Note:
1. AgBr behaves the same way and is used in photography.
2. Keep such salts in dark bottles.