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NCERT Exemplar · Class 10 Science Chemical Reactions and Equations

44 questions · 44 still being checked

Long Answer Questions 39–44 (part 5 of 5)

  1. Exercise 39

    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)
    Identity the brown gas X evolved.
    (c)
    Identity the type of reaction.
    (d)
    What could be the pH range of aqueous solution of the gas X?

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    NCERT’s answer
    (a)
    Balanced chemical equation
    \[2 \mathrm{Cu}\left(\mathrm{NO}_3\right)_2(\mathrm{~s}) \xrightarrow{\text { Heat }} 2 \mathrm{CuO}(\mathrm{~s})+\mathrm{O}_2(\mathrm{~g})+4 \mathrm{NO}_2(\mathrm{~g}) \]
    (b)
    The brown gas X evolved is nitrogen dioxide \(\displaystyle \left(\mathrm{NO}_2\right)\)
    (c)
    This is a decomposition reaction
    (d)
    Nitrogen dioxide dissolves in water to form acidic solution because it is an oxide of non-metal. Therefore, pH of this solution is less than $\displaystyle 7$
    (a)
    \[2\mathrm{Cu(NO_3)_2(s)} \xrightarrow{\Delta} 2\mathrm{CuO(s)} + 4\mathrm{NO_2(g)} + \mathrm{O_2(g)} \]
    (b)
    X is \(\displaystyle \mathrm{NO_2}\), nitrogen dioxide.
    (c)
    Thermal decomposition; also redox (N: \(\displaystyle +5\rightarrow+4\), O: \(\displaystyle -2\rightarrow0\)).
    (d)
    \(\displaystyle \mathrm{NO_2}\) dissolves in water to give an acid:
    \[3\mathrm{NO_2(g)} + \mathrm{H_2O(l)} \rightarrow 2\mathrm{HNO_3(aq)} + \mathrm{NO(g)} \]
    The solution is acidic, \(\displaystyle \mathrm{pH} < 7\).
    Answer: X is \(\displaystyle \mathrm{NO_2}\); thermal decomposition; aqueous \(\displaystyle \mathrm{NO_2}\) is acidic, \(\displaystyle \mathrm{pH} < 7\).

  2. Exercise 40

    Give the characteristic tests for the following gases
    (a)
    CO2\displaystyle \mathrm{CO}_2
    (b)
    SO2\displaystyle \mathrm{SO}_2
    (c)
    O2\displaystyle \mathrm{O}_2
    (d)
    H2\displaystyle \mathrm{H}_2

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    NCERT’s answer
    The characteristic test for
    (a)
    Carbon dioxide \(\displaystyle \left(\mathrm{CO}_2\right)\) gas turns lime water milky when passed through it due to the formation of insoluble calcium carbonate.
    \[\underset{\text{Lime water}}{\mathrm{Ca}(\mathrm{OH})_2}+\underset{\text{Carbon dioxide}}{\mathrm{CO}_2} \rightarrow \underset{\text{Calcium carbonate}}{\mathrm{CaCO}_3}+\mathrm{H}_2 \mathrm{O} \]
    (b)
    Sulphur dioxide \(\displaystyle \left(\mathrm{SO}_2\right)\) gas when passed through acidic potassium permanganate solution (purple in colour) turns it colourless because \(\displaystyle \mathrm{SO}_2\) is a strong reducing agent
    \[\underset{\text{Potasssium permanganate (Purple)}}{2 \mathrm{KMnO}_4}+2 \mathrm{H}_2 \mathrm{O}+\underset{\text{Sulphur dioxide}}{5 \mathrm{SO}_2} \rightarrow \underset{\text{Potassium sulphate (Colourless)}}{\mathrm{K}_2 \mathrm{SO}_4}+\underset{\text{Manganese sulphate (Colourless)}}{2 \mathrm{MnSO}_4}+2 \mathrm{H}_2 \mathrm{SO}_4 \]
    or Sulphur dioxide gas when passed through acidic dichromate solution (orange in colour) turns it to green because sulphur dioxide is a strong reducing agent.
    (c)
    The evolution of oxygen \(\displaystyle \left(\mathrm{O}_2\right)\) gas during a reaction can be confirmed by bringing a burning candle near the mouth of the test tube containing the reaction mixture. The intensity of the flame increases because oxygen supports burning.
    (d)
    Hydrogen \(\displaystyle \left(\mathrm{H}_2\right)\) gas burns with a pop sound when a burning candle is brought near it.
    \(\displaystyle \mathrm{CO_2}\): lime water turns milky, \[\mathrm{Ca(OH)_2(aq)} + \mathrm{CO_2(g)} \rightarrow \mathrm{CaCO_3(s)}\downarrow + \mathrm{H_2O(l)} \] excess gas clears it again, \[\mathrm{CaCO_3(s)} + \mathrm{CO_2(g)} + \mathrm{H_2O(l)} \rightarrow \mathrm{Ca(HCO_3)_2(aq)} \]\(\displaystyle \mathrm{SO_2}\): lime water also turns milky, so use acidified \(\displaystyle \mathrm{K_2Cr_2O_7}\) paper, which turns orange \(\displaystyle \rightarrow\) green, \[\mathrm{K_2Cr_2O_7(aq)} + \mathrm{H_2SO_4(aq)} + 3\mathrm{SO_2(g)} \rightarrow \mathrm{K_2SO_4(aq)} + \mathrm{Cr_2(SO_4)_3(aq)} + \mathrm{H_2O(l)} \]\(\displaystyle \mathrm{O_2}\): a glowing splinter rekindles into flame.\(\displaystyle \mathrm{H_2}\): a burning splinter brought near the gas burns with a pop, \[2\mathrm{H_2(g)} + \mathrm{O_2(g)} \rightarrow 2\mathrm{H_2O(l)} \]Caution: pass \(\displaystyle \mathrm{CO_2}\) only briefly; excess redissolves the precipitate and the test is misread as "no gas".Answer: \(\displaystyle \mathrm{CO_2}\) turns lime water milky; \(\displaystyle \mathrm{SO_2}\) turns acidified dichromate orange \(\displaystyle \rightarrow\) green; \(\displaystyle \mathrm{O_2}\) rekindles a glowing splinter; \(\displaystyle \mathrm{H_2}\) burns with a pop.

  3. Exercise 41

    What happens when a piece of
    (a)
    zinc metal is added to copper sulphate solution?
    (b)
    aluminium metal is added to dilute hydrochloric acid?
    (c)
    silver metal is added to copper sulphate solution?
    Also, write the balanced chemical equation if the reaction occurs

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    NCERT’s answer
    (a)
    Zinc being more reactive than copper displaces copper from its solution and a solution of zinc sulphate is obtained
    \[\mathrm{Zn}(\mathrm{~s})+\underset{\text{Blue}}{\mathrm{CuSO}_4(\mathrm{aq})} \rightarrow \underset{\text{Colourless}}{\mathrm{ZnSO}_4(\mathrm{aq})}+\mathrm{Cu}(\mathrm{~s}) \]
    This is an example of displacement reaction
    (b)
    Aluminium being more reactive displaces hydrogen from dilute hydrochloric acid solution and hydrogen gas is evolved.
    \[2 \mathrm{Al}(\mathrm{~s})+6 \mathrm{HCl}(\mathrm{aq}) \rightarrow \underset{\text{Aluminium chloride}}{2 \mathrm{AlCl}_3(\mathrm{aq})}+3 \mathrm{H}_2(\mathrm{~g}) \]
    (c)
    Silver metal being less reactive than copper cannot displace copper from its salt solution. Therefore, no reaction occurs
    \[\mathrm{Ag}(\mathrm{~s})+\mathrm{CuSO}_4(\mathrm{aq}) \rightarrow \text { No reaction } \]
    (a)
    \(\displaystyle \mathrm{Zn}\) is more reactive than \(\displaystyle \mathrm{Cu}\): it displaces \(\displaystyle \mathrm{Cu}\), the blue colour fades and reddish-brown \(\displaystyle \mathrm{Cu}\) deposits.
    \[\mathrm{Zn(s)} + \mathrm{CuSO_4(aq)} \rightarrow \mathrm{ZnSO_4(aq)} + \mathrm{Cu(s)} \]
    (b)
    \(\displaystyle \mathrm{Al}\) lies above \(\displaystyle \mathrm{H}\) in the reactivity series: effervescence of \(\displaystyle \mathrm{H_2}\).
    \[2\mathrm{Al(s)} + 6\mathrm{HCl(aq)} \rightarrow 2\mathrm{AlCl_3(aq)} + 3\mathrm{H_2(g)}\uparrow \]
    (c)
    \(\displaystyle \mathrm{Ag}\) is less reactive than \(\displaystyle \mathrm{Cu}\): no displacement, no reaction.
    Answer: (a) \(\displaystyle \mathrm{Cu}\) deposits, \(\displaystyle \mathrm{ZnSO_4}\) forms; (b) \(\displaystyle \mathrm{H_2}\) evolves, \(\displaystyle \mathrm{AlCl_3}\) forms; (c) no reaction.

  4. Exercise 42

    What happens when zinc granules are treated with dilute solution of H2SO4,HCl,HNO3,NaCl\displaystyle \mathrm{H}_2 \mathrm{SO}_4, \mathrm{HCl}, \mathrm{HNO}_3, \mathrm{NaCl} and NaOH, also write the chemical equations if reaction occurs.

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    NCERT’s answer
    The reaction of Zn granules with
    (a)
    Dilute \(\displaystyle \mathrm{H}_2 \mathrm{SO}_4\)
    \[\mathrm{Zn}(\mathrm{~s})+\mathrm{H}_2 \mathrm{SO}_4(\mathrm{aq}) \rightarrow \mathrm{ZnSO}_4(\mathrm{aq})+\mathrm{H}_2(\mathrm{~g}) \]
    (b)
    Dilute HCl
    \[\mathrm{Zn}(\mathrm{~s})+2 \mathrm{HCl}(\mathrm{aq}) \rightarrow \mathrm{ZnCl}_2(\mathrm{aq})+\mathrm{H}_2(\mathrm{~g}) \]
    (c)
    Dilute \(\displaystyle \mathrm{HNO}_3\)
    Reaction with dilute \(\displaystyle \mathrm{HNO}_3\) is different as compared to other acids because nitric acid is an oxidising agent and it oxidises \(\displaystyle \mathrm{H}_2\) gas evolved to \(\displaystyle \mathrm{H}_2 \mathrm{O}\).
    \[4 \mathrm{Zn}(\mathrm{~s})+10 \mathrm{HNO}_3(\mathrm{aq}) \rightarrow 4 \mathrm{Zn}\left(\mathrm{NO}_3\right)_2(\mathrm{aq})+5 \mathrm{H}_2 \mathrm{O}(\mathrm{l})+\mathrm{N}_2 \mathrm{O}(\mathrm{~g}) \]
    (d)
    NaCl solution
    \[\mathrm{Zn}(\mathrm{~s})+\mathrm{NaCl}(\mathrm{aq}) \rightarrow \text { No reaction } \]
    (e)
    NaOH solution
    \[\mathrm{Zn}(\mathrm{~s})+2 \mathrm{NaOH}(\mathrm{aq}) \rightarrow \underset{\text{Sodium zincate}}{\mathrm{Na}_2 \mathrm{ZnO}_2(\mathrm{aq})}+\mathrm{H}_2(\mathrm{~g}) \]
    \(\displaystyle \mathrm{H_2SO_4}\) (dil.): effervescence, \(\displaystyle \mathrm{H_2}\) evolves, \[\mathrm{Zn(s)} + \mathrm{H_2SO_4(aq)} \rightarrow \mathrm{ZnSO_4(aq)} + \mathrm{H_2(g)}\uparrow \]\(\displaystyle \mathrm{HCl}\) (dil.): effervescence, \(\displaystyle \mathrm{H_2}\) evolves, \[\mathrm{Zn(s)} + 2\mathrm{HCl(aq)} \rightarrow \mathrm{ZnCl_2(aq)} + \mathrm{H_2(g)}\uparrow \]\(\displaystyle \mathrm{HNO_3}\) (dil.): an oxidising acid, so no \(\displaystyle \mathrm{H_2}\); it is reduced to \(\displaystyle \mathrm{N_2O}\), \[4\mathrm{Zn(s)} + 10\mathrm{HNO_3(aq)} \rightarrow 4\mathrm{Zn(NO_3)_2(aq)} + \mathrm{N_2O(g)}\uparrow + 5\mathrm{H_2O(l)} \]\(\displaystyle \mathrm{NaCl}\): \(\displaystyle \mathrm{Zn}\) is less reactive than \(\displaystyle \mathrm{Na}\), so it cannot displace it; no reaction.\(\displaystyle \mathrm{NaOH}\): \(\displaystyle \mathrm{Zn}\) reacts with the base, forming sodium zincate and \(\displaystyle \mathrm{H_2}\), \[\mathrm{Zn(s)} + 2\mathrm{NaOH(aq)} \rightarrow \mathrm{Na_2ZnO_2(aq)} + \mathrm{H_2(g)}\uparrow \]Answer: \(\displaystyle \mathrm{H_2SO_4}\), \(\displaystyle \mathrm{HCl}\), \(\displaystyle \mathrm{NaOH}\): \(\displaystyle \mathrm{H_2}\) evolves; dil. \(\displaystyle \mathrm{HNO_3}\): \(\displaystyle \mathrm{N_2O}\) forms, no \(\displaystyle \mathrm{H_2}\); \(\displaystyle \mathrm{NaCl}\): no reaction.

  5. Exercise 43

    On adding a drop of barium chloride solution to an aqueous solution of sodium sulphite, white precipitate is obtained.
    (a)
    Write a balanced chemical equation of the reaction involved
    (b)
    What other name can be given to this precipitation reaction?
    (c)
    On adding dilute hydrochloric acid to the reaction mixture, white precipitate disappears. Why?

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    NCERT’s answer
    (a)
    Balanced chemical equation
    \[\underset{\text{Sodium sulphite}}{\mathrm{Na}_2 \mathrm{SO}_3(\mathrm{aq})}+\underset{\text{Barium chloride}}{\mathrm{BaCl}_2(\mathrm{aq})} \rightarrow \underset{\text{Barium sulphite}}{\mathrm{BaSO}_3(\mathrm{~s})}+\underset{\text{Sodium chloride}}{2 \mathrm{NaCl}(\mathrm{aq})} \]
    (b)
    This reaction is also known as double displacement reaction
    (c)
    \(\displaystyle \mathrm{BaSO}_3\) is a salt of a weak acid \(\displaystyle \left(\mathrm{H}_2 \mathrm{SO}_3\right)\), therefore dilute acid such as HCl decomposes barium sulphite to produce sulphur dioxide gas which has the smell of burning sulphur.
    \[\underset{\text{White ppt.}}{\mathrm{BaSO}_3(\mathrm{~s})}+2 \mathrm{HCl}(\mathrm{aq}) \rightarrow \mathrm{BaCl}_2+\mathrm{H}_2 \mathrm{O}+\mathrm{SO}_2(\mathrm{~g}) \]
    \(\displaystyle \mathrm{BaCl}_2\) is soluble in water, hence white precipitate disappears
    (a)
    \[\mathrm{BaCl_2(aq)} + \mathrm{Na_2SO_3(aq)} \rightarrow \mathrm{BaSO_3(s)}\downarrow + 2\mathrm{NaCl(aq)} \]
    (b)
    A double displacement reaction.
    (c)
    \(\displaystyle \mathrm{BaSO_3}\) is a salt of the weak, volatile acid \(\displaystyle \mathrm{H_2SO_3}\); the stronger \(\displaystyle \mathrm{HCl}\) displaces it, giving soluble \(\displaystyle \mathrm{BaCl_2}\) and \(\displaystyle \mathrm{SO_2}\) gas:
    \[\mathrm{BaSO_3(s)} + 2\mathrm{HCl(aq)} \rightarrow \mathrm{BaCl_2(aq)} + \mathrm{SO_2(g)}\uparrow + \mathrm{H_2O(l)} \]
    Answer: \(\displaystyle \mathrm{BaSO_3}\) is precipitated (double displacement); it reacts with \(\displaystyle \mathrm{HCl}\) to give soluble \(\displaystyle \mathrm{BaCl_2}\) and \(\displaystyle \mathrm{SO_2}\) gas, so the precipitate disappears.

  6. Exercise 44

    You are provided with two containers made up of copper and aluminium. You are also provided with solutions of dilute HCl, dilute HNO3,ZnCl2\displaystyle \mathrm{HNO}_3, \mathrm{ZnCl}_2 and H2O\displaystyle \mathrm{H}_2 \mathrm{O}. In which of the above containers these solutions can be kept?

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    NCERT’s printed answer for this exercise does not match its own question. This working follows the question as printed.

    Reactivity series (relevant part): \(\displaystyle \mathrm{Al > Zn > H > Cu}\).Copper container: \(\displaystyle \mathrm{Cu}\) lies below \(\displaystyle \mathrm{H}\) and \(\displaystyle \mathrm{Zn}\), so dilute \(\displaystyle \mathrm{HCl}\), \(\displaystyle \mathrm{ZnCl_2}\) and \(\displaystyle \mathrm{H_2O}\) do not react (safe). Dilute \(\displaystyle \mathrm{HNO_3}\) is an oxidising acid and attacks \(\displaystyle \mathrm{Cu}\) (unsafe): \[3\mathrm{Cu(s)} + 8\mathrm{HNO_3(aq)} \rightarrow 3\mathrm{Cu(NO_3)_2(aq)} + 2\mathrm{NO(g)}\uparrow + 4\mathrm{H_2O(l)} \]Aluminium container: \(\displaystyle \mathrm{Al}\) lies above \(\displaystyle \mathrm{H}\) and \(\displaystyle \mathrm{Zn}\), so dilute \(\displaystyle \mathrm{HCl}\) and \(\displaystyle \mathrm{ZnCl_2}\) both react (unsafe): \[2\mathrm{Al(s)} + 6\mathrm{HCl(aq)} \rightarrow 2\mathrm{AlCl_3(aq)} + 3\mathrm{H_2(g)}\uparrow \] \[2\mathrm{Al(s)} + 3\mathrm{ZnCl_2(aq)} \rightarrow 2\mathrm{AlCl_3(aq)} + 3\mathrm{Zn(s)} \] Dilute \(\displaystyle \mathrm{HNO_3}\) oxidises \(\displaystyle \mathrm{Al}\) too (unsafe): \[\mathrm{Al(s)} + 4\mathrm{HNO_3(aq)} \rightarrow \mathrm{Al(NO_3)_3(aq)} + \mathrm{NO(g)}\uparrow + 2\mathrm{H_2O(l)} \] \(\displaystyle \mathrm{H_2O}\) is safe: the oxide film protects \(\displaystyle \mathrm{Al}\).Caution: only concentrated \(\displaystyle \mathrm{HNO_3}\) makes \(\displaystyle \mathrm{Al}\) passive; dilute \(\displaystyle \mathrm{HNO_3}\) does not.Answer: \(\displaystyle \mathrm{Cu}\) container: dil. \(\displaystyle \mathrm{HCl}\), \(\displaystyle \mathrm{ZnCl_2}\), \(\displaystyle \mathrm{H_2O}\). \(\displaystyle \mathrm{Al}\) container: \(\displaystyle \mathrm{H_2O}\) only.