Chapter 1 · Chemical Reactions and Equations

NCERT Exemplar Class 10 Science Chapter 1 Long Answer Questions

All 6 questions of NCERT Exemplar Class 10 Science Chapter 1 Long Answer Questions (Chemical Reactions and Equations), in the book's order, each with its answer and a step-by-step solution.

1. Q39
   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

   1. (a)Write a balanced ​chemical ​equation of the reaction.
   2. (b)Identity the brown gas X evolved.
   3. (c)Identity the ​type of reaction.
   4. (d)What could be the pH range of aqueous solution of the gas X?
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   Gas X is ${\text{NO}}_2$, formed by thermal decomposition of copper(II) nitrate; its aqueous solution is acidic, with pH below $7$.

   (a) $2{\text{Cu(NO}}_3{)}_2(s)\stackrel{\text{Heat}}{\to }2\text{CuO}(s)+4{\text{NO}}_2(g)+{\text{O}}_2(g)$; (b) The brown gas X is nitrogen dioxide, ${\text{NO}}_2$ — the only brown gas the elements present can give; (c) It is a decomposition reaction, specifically a thermal decomposition, and it is endothermic; within the nitrate ion it is also a redox change, since nitrogen falls from $+5$ to $+4$ while some oxygen rises from $-2$ to $0$; (d) Nitrogen dioxide dissolves in water to give nitrous and nitric acid, $2{\text{NO}}_2+{\text{H}}_2\text{O}\to {\text{HNO}}_2+{\text{HNO}}_3$, so the solution is acidic and its pH lies below $7$, in the range $0$ to $7$ — and because nitric acid is a strong acid the value falls to the low single figures.

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2. Q40
   Give ​the ​characteristic ​tests for ​the following ​gases

   1. (a)${\text{CO}}_2$
   2. (b)${\text{SO}}_2$
   3. (c)${\text{O}}_2$
   4. (d)${\text{H}}_2$
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   ${\text{CO}}_2$ turns lime water milky; ${\text{SO}}_2$ turns acidified dichromate paper orange to green; ${\text{O}}_2$ relights a glowing splint; ${\text{H}}_2$ burns with a pop.

   (a) ${\text{CO}}_2$: pass the gas through freshly prepared lime water — it turns milky, because insoluble calcium carbonate is formed, ${\text{Ca(OH)}}_2(aq)+{\text{CO}}_2(g)\to {\text{CaCO}}_3(s)+{\text{H}}_2\text{O}(l)$ (the milkiness disappears if excess gas is passed, as the soluble hydrogencarbonate forms); (b) ${\text{SO}}_2$: pass the gas over paper moistened with acidified potassium dichromate — the orange colour turns green, because sulphur dioxide is a reducing agent and reduces the dichromate to a chromium(III) salt; the gas also smells of burning sulphur and turns moist blue litmus red; (c) ${\text{O}}_2$: bring a glowing splint to the mouth of the tube — it relights and burns brightly, because oxygen supports combustion without itself burning; (d) ${\text{H}}_2$: bring a burning splint to the mouth of the tube — the gas burns with a pop sound, $2{\text{H}}_2(g)+{\text{O}}_2(g)\to 2{\text{H}}_2\text{O}(l)$, and the flame is not supported further, because hydrogen burns but does not support combustion.

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3. Q41
   What happens when a piece of

   1. (a)zinc metal is added to copper sulphate ​solution?
   2. (b)​aluminium metal is added to ​dilute hydrochloric ​acid?
   3. (c)​silver metal ​is added to copper ​sulphate solution? Also, write the balanced chemical equation if the reaction occurs
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   Zinc displaces copper, aluminium displaces hydrogen from the acid, and silver does not react with copper sulphate.

   (a) Zinc lies above copper in the reactivity series, so it displaces copper from the solution: the blue colour of the copper sulphate slowly fades, a reddish-brown deposit of copper forms on the zinc, and the mixture warms slightly. $\text{Zn}(s)+{\text{CuSO}}_4(aq)\to {\text{ZnSO}}_4(aq)+\text{Cu}(s)$; (b) Aluminium lies well above hydrogen, so it displaces hydrogen from the acid: there is brisk effervescence of hydrogen gas, the metal dissolves and the mixture becomes hot. $2\text{Al}(s)+6\text{HCl}(aq)\to 2{\text{AlCl}}_3(aq)+3{\text{H}}_2(g)$; (c) Silver lies below copper in the reactivity series, so it cannot displace copper. No reaction takes place — the blue colour is unchanged and the silver is unaffected, so no equation can be written.

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4. Q42
   What ​happens when ​zinc granules are treated with dilute ​solution of ${\text{H}}_2{\text{SO}}_4$, $\text{HCl}$, ${\text{HNO}}_3$, $\text{NaCl}$ ​and $\text{NaOH}$, ​also ​write the ​chemical equations if reaction occurs.
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   Zinc gives hydrogen with dilute ${\text{H}}_2{\text{SO}}_4$, $\text{HCl}$ and $\text{NaOH}$, an oxide of nitrogen with ${\text{HNO}}_3$, and no reaction with $\text{NaCl}$.

   Dilute ${\text{H}}_2{\text{SO}}_4$: zinc is above hydrogen, so hydrogen is evolved and zinc sulphate is formed — $\text{Zn}(s)+{\text{H}}_2{\text{SO}}_4(aq)\to {\text{ZnSO}}_4(aq)+{\text{H}}_2(g)$. Dilute $\text{HCl}$: the same, giving zinc chloride — $\text{Zn}(s)+2\text{HCl}(aq)\to {\text{ZnCl}}_2(aq)+{\text{H}}_2(g)$. Dilute ${\text{HNO}}_3$: the zinc dissolves, but no hydrogen is evolved, because nitric acid is a strong oxidising agent and oxidises the hydrogen produced to water while its own nitrogen is reduced to an oxide of nitrogen — with very dilute acid, $4\text{Zn}(s)+10{\text{HNO}}_3(aq)\to 4{\text{Zn(NO}}_3{)}_2(aq)+{\text{N}}_2\text{O}(g)+5{\text{H}}_2\text{O}(l)$, and with dilute acid the reduction product is nitric oxide, $3\text{Zn}(s)+8{\text{HNO}}_3(aq)\to 3{\text{Zn(NO}}_3{)}_2(aq)+2\text{NO}(g)+4{\text{H}}_2\text{O}(l)$. Dilute $\text{NaCl}$: no reaction, because sodium chloride is a neutral salt and sodium is far more reactive than zinc, so zinc cannot displace it. Dilute $\text{NaOH}$: hydrogen is evolved, because zinc is amphoteric and reacts with alkalis as well as with acids, dissolving as sodium zincate — $\text{Zn}(s)+2\text{NaOH}(aq)\to {\text{Na}}_2{\text{ZnO}}_2(aq)+{\text{H}}_2(g)$.

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5. Q43
   On adding a drop of barium chloride solution to an aqueous solution of sodium sulphite, white precipitate is obtained.

   1. (a)​Write a balanced ​chemical equation of the reaction involved
   2. (b)What other name can ​be given to ​this precipitation reaction?
   3. (c)On ​adding dilute hydrochloric acid ​to the reaction ​mixture, white ​precipitate disappears. Why?
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   Barium sulphite precipitates in a double displacement reaction; hydrochloric acid dissolves it, releasing ${\text{SO}}_2$ gas.

   (a) ${\text{BaCl}}_2(aq)+{\text{Na}}_2{\text{SO}}_3(aq)\to {\text{BaSO}}_3(s)+2\text{NaCl}(aq)$ — the white precipitate is barium sulphite, ${\text{BaSO}}_3$; (b) It can also be called a double displacement reaction, since the two compounds simply exchange their ions; (c) Sulphite is the anion of the weak acid ${\text{H}}_2{\text{SO}}_3$, so the much stronger hydrochloric acid takes it out of the solid, forming soluble barium chloride and sulphurous acid. The sulphurous acid at once breaks up into water and sulphur dioxide, which escapes as a gas with the smell of burning sulphur, so the change cannot reverse and the white precipitate dissolves completely: ${\text{BaSO}}_3(s)+2\text{HCl}(aq)\to {\text{BaCl}}_2(aq)+{\text{H}}_2\text{O}(l)+{\text{SO}}_2(g)$.

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6. Q44
   You are ​provided with two containers made ​up of copper ​and aluminium. You are also provided ​with solutions of dilute $\text{HCl}$, dilute ${\text{HNO}}_3$, ${\text{ZnCl}}_2$ and ${\text{H}}_2\text{O}$. In ​which of the ​above containers ​these solutions can be kept?
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   The copper container can hold dilute $\text{HCl}$, ${\text{ZnCl}}_2$ and water but not dilute ${\text{HNO}}_3$; the aluminium container can hold only water.

   In the copper container: dilute $\text{HCl}$, ${\text{ZnCl}}_2$ solution and ${\text{H}}_2\text{O}$ can be kept. Copper lies below hydrogen, so dilute hydrochloric acid does not attack it; copper lies below zinc, so it cannot displace zinc from zinc chloride; and copper does not react with water. Dilute ${\text{HNO}}_3$ cannot be kept in it, because nitric acid is an oxidising acid and dissolves copper even though copper is below hydrogen: $3\text{Cu}(s)+8{\text{HNO}}_3(aq)\to 3{\text{Cu(NO}}_3{)}_2(aq)+2\text{NO}(g)+4{\text{H}}_2\text{O}(l)$. In the aluminium container: only ${\text{H}}_2\text{O}$ can be kept, since aluminium is protected from cold water by its oxide film. Dilute $\text{HCl}$ attacks it because aluminium lies above hydrogen, $2\text{Al}(s)+6\text{HCl}(aq)\to 2{\text{AlCl}}_3(aq)+3{\text{H}}_2(g)$; dilute ${\text{HNO}}_3$ attacks it as well; and ${\text{ZnCl}}_2$ solution attacks it because aluminium lies above zinc and displaces it, $2\text{Al}(s)+3{\text{ZnCl}}_2(aq)\to 2{\text{AlCl}}_3(aq)+3\text{Zn}(s)$.

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