New Simplified Chemistry Class 10 ICSE Solutions – Study Of Compounds – Sulphuric Acid
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QUESTIONS
2000
Question 1.
What do you see when concentrated sulphuric acid is added to copper sulphate 5-water.
Answer:
The colour of blue crystal of CuSO4.5H2O changes to white amorphous as the compound loses its water of crystallisation.
Question 2.
Name one catalyst used industrially which speeds up the conversion of SO2 to SO3 in the production of sulphuric acid in the laboratory or industrially. Write the equation for the conversion of sulphur dioxide to sulphur trioxide. Why does this reaction supply energy. What is the name of the compound formed between SO4 and sulphuric acid.
Answer:
V2O5; It is exothermic reaction ; oleum.
2001
Question 1.
Write equations for:
- H2SO4 – producing H2,
- Between Pb(NO3)2 and dil. H2SO4.
Answer:
- Zn + H2SO4(dil.) → ZnSO4 + H2
- Pb (NO3)2 + H2SO4 (dil.) → PbSO4 ↓+ 2HNO3
Question 2.
Explain how a reagent chosen from: ammonium hydroxide, barium chloride, sodium chloride, sodium hydroxide, sulphuric acid and nitric acid enables to distinguish between the two acids mentioned there in.
Answer:
Barium chloride can be used to distinguish between sulphuric acid and nitric acid. Out of these two acids only sulphuric acid gives a white precipitate with barium chloride solution.
2002
Question 1.
State the substance/s reacted with dilute or concentrated sulphuric acid to form the following gases:
- Hydrogen
- Carbon dioxide.
State whether the acid used in each case is dilute or concentrated.
Answer:
(1) Zinc (or any other reactive metal) reacts with dil. H2SO4 to give hydrogen.
Zn (s) + H2SO4 (aq.) → ZnSO4 (aq.) + H2(g)
(2) Sodium carbonate reacts with dil. H2S04 to give C02
Na2C03(s) + H2SO4(aq.) → Na2SO4(aq.) + H2O(l) + CO2(g)
The above reaction can also be carried out with NaHCO3 (sodium bicarbonate) or KHCO3 (potassium bicarbonate)
Question 2.
Write the equations for the laboratory preparation of:
- Sodium sulphate (Na2SO4) using dil. H2SO4,
- Lead sulphate (PbSO4) using dil. H2SO4.
Answer:
- 2NaOH + H2SO4 (dil) → Na2SO4 + 2H2O
- Pb(NO3)2 + H2SO4 (dil) → PbSO4 + 2HNO3
2003
Question 1.
State the name of the process by which H2SO4 is manufactured. Name the catalyst used.
Answer:
By Contact process — vanadium pentoxide (V2O5)
Question 2.
“Concentrated sulphuric acid is used in the laboratory preparation of nitric acid and hydrochloric acid because it is………. (less volatile / stronger) in comparison to these two acids.”
Answer:
Less volatile
Question 3.
Write the equations for the laboratory preparation of the following salts using sulphuric acid:
- Copper sulphate from copper
- Lead sulphate from lead nitrate
Answer:
- Cu + 2H2SO4 (dil) → CuSO4 + SO2 + H2O
- Pb (NO3)2 + H2SO4 → PbSO4 + 2HNO3
2004
Question 1.
Name the catalyst which helps in the conversion of sulphur dioxide to sulphur trioxide.
Answer:
Platinum or Vanadium pentoxide.
Question 2.
In the Contact process, sulphur trioxide is not converted to sulphuric acid by reacting it with water. Instead a two- step procedure is used. Write the equations for the two steps involved.
Answer:
The equations for the two steps involved are:
2005
Question 1.
Write balanced equations for the following:
- Potassium hydrogen carbonate and dilute sulphuric acid.
- Sodium nitrate and concentrated sulphuric acid.
Answer:
- 2KHCO. + H2SO4 (dil) → K2SO4 + 2H2O + 2CO2
- 2NaN03 + H2S04 (cone) → Na2S04 + 2HNO3
Question 2.
Choose the property of sulphuric acid (A, B, C or D), which is relevant to each of the preparations
(1) to (2) : A: dil. acid (typical acid properties), B: Non-volatile acid, C: Oxidizing agent, D: Dehydrating agent. Preparation of
(1) HCl
(2) ethene from ethanol
(3) copper sulphate from copper oxide.
Answer:
- Non volatile acid (B)
- Dehydrating agent (D)
- dil. acid (A)
2006
Question 1.
Name the process used for the large scale manufacture of sulphuric acid.
Answer:
Contact process.
Question 2.
Which property of sulphuric acid accounts for its use as a dehydrating agent.
Answer:
Sulphuric acid removes water of crystallization.
Question 3.
H2SO4 is an oxidizing agent and a non volatile acid. Write an equation for each property.
Answer:
- Sulphuric acid as an Oxidising agent —
C + 2H2 SO4 → CO2 + 2SO2 + 2H2O - Sulphuric acid as an Non-volatile acid —
Question 4.
Select the correct compound from the list — Ammonia, Copper oxide, Copper sulphate, Hydrogen chloride, Hydrogen sulphide, Lead bromide — This compound smells of rotten eggs.
Answer:
Hydrogen sulphide.
2007
Question 1.
Write balanced equation for the following reactions:
- Lead sulphate from lead nitrate solution and dilute sulphuric acid.
- Copper sulphate from copper and cone, sulphuric acid.
Answer:
- Pb(NO3)2 + H2SO4 (dil) → PbSO4 + 2HNO3
- Cu + 2H2SO4→ CuSO4 + SO2 + 2H2O
(Cone)
Question 2.
Properties of H2SO4 are listed below. Choose the property A, B, C or D which is responsible for the reactions (i) to (v). A : Acid B: Dehydrating agent C: Nonvolatile acid D: Oxidizing agent
- C12H12O11+ nH2SO4 → 12C + 11H20 + nH2SO4,
- S + 2H2SO4 → 3SO2 + 2H2O,
- NaCl + H2SO4 → NaHSO4 + HCl,
- CuO + H2SO4 → CuSO4 + H2O
- Na2CO3 + H2SO4 Na2SO4 + H2O + CO2
(Some properties may be repeated)
Answer:
(1) B (2) D (3) C (4) A (5) A
Question 3.
Dilute hydrochloric acid and dilute sulphuric acid are both colourless solutions. How will the addition of barium chloride solution to each help to distinguish between the two.
Answer:
Out of dilute hydrochloric acid and dilute sulphuric acid, Dilute hydrochloric acid will give a white ppt. of barium sulphate(BaSO4) with barium chloride solution.
H2S04 (dil.) + BaCl2 (aq.) → BaSO4 (s) + 2HCl
HCl(aq.) + BaCl2 (aq.) → No reaction
Question 4.
From HCl, HNOs, H2SO4, state which has the highest boiling point and which has the lowest.
Answer:
H2SO4 [358°C] has highest boiling point. HCl [-85°C] has lowest boiling point.
2008
Question 1.
Dilute sulphuric acid will produce a white precipitate when added to a solution of:
A.Copper nitrate
B. Zinc nitrate
C. Lead nitrate
D. Sodium nitrate
Question 2.
Identify the following substances :Liquid E can be dehydrated to product ethene.
Answer:
C2H5OH (Ethanol)
Question 3.
Copy and complete the following table relating to an important industrial processes and its final. Output refers to the product of the process not the intermediate steps.
Answer:
Question 4.
Making use only of substances given: dil. sulphuric acid, Sodium carbonate, Zinc, Sodium sulphite, Lead, Calcium carbonate: Give eqautions for the reactions by which you could obtain:
(1) hydrogen
(2) sulphur dioxide
(3) carbon dioxide
(4) zinc carbonate (2 steps)
Answer:
Question 5.
What property of cone. H2SO4 is used in the action when sugar turns black in its presence.
Answer:
Cone. Sulphuric acid is a dehydrating agent.
Question 6.
Write the equations for:
(1) dil. H2SO4 and barium chloride.
(2) dil. H2SO4 and sodium sulphide.
Answer:
Question 7.
Which property of cone. H2SO4 allows it to be used in the preparation of HCl and HNO3
Answer:
Non volatile acid.
2009
Question 1.
Name the gas evolved (formula is not acceptable). – The gas that can be oxidised to sulphur.
Answer:
Hydrogen sulphide (H2S)
2010
Question 1.
Give the equation for:
- Heat on sulphur with cone. H2SO4.
- Reaction of – sugar with cone. H2SO4.
Answer:
- Reaction of sulphur with cone. H2SO4
S + 2H2SO4 (cone.) → 3SO2 + 2H2O - Reaction of sugar with cone. H2SO4
Question 2.
Give a balanced equation for the conversion of zinc oxide to zinc sulphate.
Answer:
Question 3.
Select from A, B, C –
A: Sodium hydroxide solution.
B: A weak acid.
C: Dilute sulphuric acid.
The solution which liberates sulphur dioxide gas, from sodium sulphite.
2011
Question 1.
State your observation when – Sugar crystals are added to cone, sulphuric acid.
Answer:
A lot of effervescence takes place in the test tube. The test tube gets very hot. So in the end sugar crystals change in the black residue.
Question 2.
Choose the correct answer from the choices – The gas evolved when dil. sulphuric acid reacts with iron sulphide.
(a) Hydrogen sulphide
(b) Sulphur dioxide
(c) Sulphur trioxide
(d) Vapour of sulphuric acid.
Question 3.
Give a balanced equation for – Dilute sulphuric acid is poured over sodium sulphite
Answer:
Na2SO3 + H2SO4 (dil.) → Na2SO4 + H2O+ SO2 ↑
Question 4.
With the help of balanced equations, outline the manufacture of sulphuric acid by the contact process.
Answer:
Contact process: Sulphur or Pyrite Burner
S + O2 → SO2
Contact Tower
Absorption Tower
Dilution Tank
H2S2O7 + H2O → 2 H2SO4
Question 5.
State the property of sulphuric acid shown by the reaction of cone, sulphuric acid when heated with
(a) Potassium nitrate
(b) Carbon?
Answer:
(a) It behaves as a non volatile acid and helps in the production of a volatile acid.
(b) It behaves as an oxidising agent and oxidises carbon to carbon dioxide
C + 2H2 SO4→ CO2 ↑+ 2H2O + 2SO2 ↑
2012
Question 1.
Name – The gas produced on reaction of dilute sulphuric acid with a metallic sulphide.
Answer:
Hydrogen sulphide (H2S)
Question 2.
Some properties of sulphuric acid are listed below. Choose the role played by sulphuric acid – A, B, C, or D which is responsible for the reactions (i) to (v). Some role/s may be repeated.
A. Dilute acid
B. Dehydrating agent
C. Non-volatile acid
D. Oxidising agent
(1)
(2) S+H2SO4 (conc) →3SO2 + 2H2O
(3)
(4) MgO + H2SO4 → MgSO4 + H20
(5) Zn + 2H2SO4 (cone.) → ZnSO4 + SO2 + 2H2O
Answer:
- B: Dehydrating agent
- D: Oxidising agent
- C: Non-volatile acid
- A: Dilute acid
- D: Oxidising agent
Question 3.
Give balanced equation for the reaction : Zinc sulphide and dilute sulphuric acid.
Answer:
ZnS + H2 SO4 (dil.) → ZnSO4 + H2S
2013
Question 1.
State one appropriate observation for : Cone. H2SO4 is added to a crystal of hydrated copper sulphate.
Answer:
The blue coloured hydrated copper sulphate crystals disintegrate with a hissing sound, giving off steam and leaving behind white residue.
Question 2.
In the given equation S + 2H2SO4 → 3S02 + 2H2O: Identify the role played by cone. H2S04 i.e.
(A) Non-volatile acid
(B) Oxidising agent
(C) Dehydrating agent
(D) None of the above.
Question 3.
Give a balanced equation for : Dehydration of concentrated sulphuric acid with sugar crystals.
Answer:
Question 4.
Identify the substance underlined: A dilute mineral acid which forms a white precipitate when treated with barium chloride solution.
Answer:.
Dilute sulphuric acid.
2014
Question 1.
Write balanced equations for the following: Action of concentrated sulphuric acid on carbon.
Answer:
Action of concentrated sulphuric acid on carbon.
C + 2H2SO → CO2 + 2H20 + 2SO2
Question 2.
Distinguish between the following pairs of compounds using the test given within brackets:Dilute sulphuric acid and dilute hydrochloric acid (using barium chloride solution)
Answer:
Out of dilute H2SO4 and dilute HCl, only dilute H2SO4 gives white ppt. of BaSO4 with barium chloride solution
BaCl2 + HCl → No ppt. formation
Question 3.
State – Any two conditions for the conversion of sulphur dioxide to sulphur trioxide.
Answer:
- Two condition for the conversion of SO2 to SO3
- The mixture of SO2 gas and O2 gas must be pure and dry and in the ratio of 2 : 1 by volume.
- The mixture should be passed over platinised asbestos or vanadium pentaoxide maintained
at 450 ° C.
Question 4.
Give one equation each to show the following properties of sulphuric acid:
(1) Dehydrating property.
(2) Acidic nature.
(3) As a non-volatile acid.
Answer:
- Dehydrating property.
- Acidic nature.
CuO + H2S04 → CuSO + H2O - As a non-volatile acid.
2015
Question 1.
Identify the acid in each case:
- The acid which is used in the preparation of a nonvolatile acid.
- The acid which produces sugar charcoal from sugar.
- The acid on mixing with lead nitrate soln. produces a white ppt. which is insoluble even on heating.
Answer:
- Nitric acid (cone.)
- Cone, sulphuric acid
- Dilute hydrochloric acid
Question 2.
Give equations for the action of sulphuric acid on —
(a) Potassium hydrogen carbonate.
(b) Sulphar
Answer:
(a) Action of sulphuric acid on potassium hydrogen carbonate
2KHCO3 + H2SO4 → K2SO4 + 2H,O + 2CO2 ↑
(b) Action of sulphuric acid on sulphur
S + 2H2SO4 (cone.) → 3SO2, + 2H2O
Question 3.
In the manufacture of sulphuric acid by the Contact process, give the equations for the conversion of sulphur trioxide to sulphuric acid.
Answer:
In the contact process for the manufacture of sulphuric acid, the equations for the conversion of sulphur trioxide to sulphuric acid are
H2S2O7 + H2O → 2H2SO4
2016
Question 1.
Write balanced chemical equations for: Action of dilute Sulphuric acid on Sodium Sulphite.
Answer:
Na2SO3 + H2SO4(dil) → Na2SO4 + H2O + SO2↑
Question 2.
State your observations when:
- Barium chloride soln. is mixed with sodium sulphate soln.
- Concentrated sulphuric acid is added to sugar crystals.
Answer:
(1)
When sodium sulphate is mixed with barium chloride. White coloured precipitates of Barium sulphate are formed.
(2)
When cone, sulphuric acid is added to sugar crystals black spongy mass (sugar charcoal) is formed.
Question 3.
A, B, C and D summarize the properties of sulphuric acid depending on whether it is dilute or concentrated.
A: Typical acid property
B: Non-volatile acid
C: Oxidizing agent
D: Dehydrating agent
Choose the property (A, B, C or D) depending on which is relevant to each of the following:
- Preparation of hydrogen chloride gas.
- Preparation of copper sulphate from copper oxide.
- Action of cone, sulphuric acid on sulphur.
Answer:
- Preparation of Hydrogen chloride gas.
B: Non-volatile acid - Preparation of Copper sulphate from copper oxide.
A: Typical acid property - Action of cone. Sulphuric acid on Sulphur.
C: Oxidizing agent
2017
Question 1.
Write the balanced chemical equation for – Action of concentrated sulphuric acid on sulphur.
Answer:
Question 2.
State one relevant observation for – Action of cone, sulphuric acid on hydrated copper sulphate.
Answer:
Blue coloured copper sulphate crystals crumble with a hissing sound and change to white powdery mass.
Question 3.
State – How will you distinguish between dilute hydrochloric acid and dilute sulphuric acid using lead nitrate solution.
Answer:
Hydrochloric acid forms a white precipitate with lead nitrate solution. This precipitate dissolves on warming the reaction mixture so as to form clear solution. Sulphuric acid forms a white precipitate with lead nitrate solution. This precipitate does not dissolve on warming the reaction mixture.
Question 4.
Write balanced chemical equations to show –
- The oxidizing action of cone, sulphuric acid on carbon.
- The behaviour of H2SO4 as an acid when it reacts with magnesium.
- The dehydrating property of cone, sulphuric acid with sugar.
- The conversion of SO3 to sulphuric acid in the Contact process.
Answer:
- C + 2H2SO4 (cone.) → CO2 + 2SO2 + 2H2O
- Mg + H2SO4 (dil.) → MgSO4 + H2 (g)
- C12H22O11 + 11 H2S04(conc.) → 12C + 11H2S04.H2O + ΔH
- (a) SO3+ H2SO4 (cone.) → H2S2O7 (oleum)
(b) H2S2O7 + H2O → 2H2SO4 (cone.)
ADDITIONAL QUESTIONS
Question 1.
State why sulphuric acid was called – ‘oil of vitriol’.
Answer:
Sulphuric acid was initially called ‘oil of vitriol ’.It was initially prepared by – distilling green vitriol [FeSO4.7H2O] and hence the name – ‘oil of vitriol’.
Question 2.
State how you would convert
(1) sulphur
(2) chlorine
(3) sulphur dioxide – to sulphur acid.
Answer:
- S + 6HNO. [cone.] → 6NO2 + 2H2O + H2SO4
- Cl2 + SO2 + 2H2O → 2HCl + H2SO4
- 3SO2 + 2HNO3 + 2H2O → 3H2SO4 + 2NO
Question 3.
State the purpose of the ‘Contact process’.
Answer:
When sulphur is burnt in air, it bums with a pale blue flame forming sulphur dioxide and traces of sulphur trioxide.
S + O2→ SO2 [2S + 3O2 → 2SO4 (traces)]
Burning of sulphur or iron pyrites in oxygen is preferred to purified air since heat energy is wasted in heating the unreactive nitrogen component of the air.
Question 4.
In the Contact process
- State how you would convert (a) sulphur (b) iron pyrites to sulphur dioxide in the first step of the Contact process.
- State the conditions i.e. catalyst, promoter, temperature and pressure in the catalytic oxidation of sulphur dioxide to sulphur trioxide in the Contact tower. Give a balanced equation for the same.
- State why the above catalytic oxidation {reaction supplies energy.
- Give a reason why – vanadium pentoxide is preferred to platinum during the catalytic oxidation of sulphur dioxide.
- Give a reason why the catalyst mass is heated electrically – only initially.
- State why sulphur trioxide vapours are absorbed in concentrated sulphuric acid and not in water to obtain sulphuric acid.
Answer:
- S + O2 → 5- SO2
4FeS2 + 11O2 → 5- 2Fe2O3 + 8SO2
- Catalytic oxidation of sulphur dioxide to sulphur trioxide.
[Above equation for the catalysed reaction is exothermic- hence supplies energy.]
Catalyst: Vanadium pentoxide [V2O5] or platinum [Pt],
Temperature: 450-500°C Pressure : 1 to 2 atmospheres
Conversion ratio: 98% of sulphur dioxide converted to sulphur trioxide. - The catalytic oxidation of SO2 to SO3 is an – exothermic reaction. Thus this reaction supplies energy in the form of heat.
- Vanadium pentoxide is preferred – to platinized asbestos as a catalyst since it is comparatively – cheaper and less easily poisoned or susceptible to impurities.
- The catalyst – mass is – only initially heated electrically, since the catalytic oxidation of sulphur dioxide is an exothermic reaction and the heat produced maintains the temperature at 450 – 500°C.
- Even though sulphur trioxide is an acid anhydride of sulphuric acid – it is not directly absorbed in water to give sulphuric acid.
The reaction is highly exothermic resulting in production of – a dense fog of sulphuric acid particles which do not condense easily.
Hence sulphur trioxide vapours are – dissolved in cone, sulphuric acid to give oleum which on dilution with – the requisite amount of soft water in the dilution tank gives – sulphuric acid of the desired concentration [about 98%].
Question 5.
Give a reason why concentrated sulphuric acid is kept in air tight bottles.
Answer:
Concentrated sulphuric acid has a great affinity for water and as such it is a hygroscopic liquid. Being Hygroscopic, it absorbs moisture from the atmosphere and hence cone, sulphuric acid is kept in air tight bottles.
Question 6.
State the basic steps with reasons, involved in diluting a beaker of cone. H2SO4.
Answer:
For dilution of cone. H2SO4, the cone, acid is always added to water and water is never added to the cone, acid even though heat is evolved in both cases.
Reason: If water is added to cone. H2SO4 the heat produced is sufficient to spontaneously vaporise a part of the few drops of water added. This is because the amount of water is very small and bioling point of water is much lower than cone. H2SO4, which is in bulk. Due to this sudden vaporisation of water cone, acid tend to spurt out and cause serious injuries.
On the other hand, if cone. H2SO4 is added to water, the heat produced in this case can only raise the temperature of water slightly because water is an bulk. Thus, in this case spurting of the cone, acid is avoided.
Question 7.
Give reasons why dilute sulphuric acid:
(1) behaves as an acid when dilute.
(2) is dibasic in nature.
Answer:
- Acidic properties of sulphuric acid are due to the presence of – hydronium ions [H3O+ ] formed when H2SO4 dissociates in aq. solution.
- Sulphuric acid dissociates in – aq. solution giving 2H+ ions per molecule – of the acids .Hence its – basically is two.
[Basicity is the number of H+ ions formed by dissociation of one molecule of the acid in its aq. soln.]
Question 8.
Convert dil. H.SO, to –
- Hydrogen
- Carbon dioxide
- Sulphur dioxide
- Hydrogen sulphide
- An acid salt
- A normal salt.
Answer:
- Dil H2SO4 to hydrogen by the action of any active metal (say zinc) on dil. H2SO4
- Dil. H2SO4t0 carbon dioxide by the action of any carbonate or bicarbonate on dil. H2SO4
- Dil. H2S04to sulphur dioxide by the action of any sulphide on dil. H2SO4
Na2SO3 + H2SO4 → Na2SO4 + H2O + SO2
- H2SO4 to hydrogen sulphide by the action of any sulphide on dil. H2SO4
FeS + H2SO4 (dil.) → FeSO4 + H2S↑ - H2S04to an acid by the action of insufficient strong base with excess of dil. H2SO4
- H2SO4to a normal salt by the action of sufficient (or excess of) strong base (NaOH) with excess of dil. FI2SO4
Question 9.
Give equations for formation of two different acids from cone. H2SO4. State the property of sulphuric acid involved in the above formation.
Answer:
Property of Sulphuric acid involved in the formation of these acids: Cone. H2SO4 is a non-volatile acid.
Question 10.
Give equations for oxidation of cone. H2SO4 giving the oxidised products –
- Carbon dioxide
- Sulphur dioxide
- Phosphoric acid
- Copper (II) sulphate
- Iodine
- Sulphur respectively.
Answer:
Question 11.
Give a reason why concentrated and not dil H2SO4 – behave as an oxidising and dehydrating agent.
Answer:
- Cone. H2SO4 be aves as an oxidising agent because cone.H2SO4 whes h. ated decomposes to give nascent oxygen which acts as a strong oxidising agent.
On the other hand, dil H2SO4 on heating does not decompose to give nascent oxygen and as such cannot behave as an oxidising agent. - Cone. H2SO4 behaves as a dehydrating agent because cone. H2SO4 act as a great affinity for water and hydration of cone. H2SO4 is an exothermic reaction.On the other hand, dil H2SO4 has no affinity for water and hence cannot act as a hydrating agent.
Question 12.
Give the equation for the reaction cone, sulphuric acid with –
- glucose
- sucrose
- cellulose
- an organic acid containing one carbon atom and two hydrogen atoms
- an organic acid containing two carbon and two hydrogen atoms
- an alcohol
- hydrated copper (II) sulphate.
Answer:
- Reaction of cone. H2SO4 with glucose, CtH12O6 –
Dehydration.
- Reaction of cone. H2SO4 with sucrose, C12H12O12 –
Dehydration.
- Reaction of cone. H2SO4 with cellulose, (C6H10O5)n –
Dehydration.
- Reaction of cone. H2SO4 with an organic acid containing one carbon atom and two hydrogen atoms, HCOOH (formic acid or methanoic acid) –
Dehydration.
- Reaction of cone. H2SO4 with an organic acid containing two carbon atom and two hydrogen atoms, [COOH]2 (oxalic acid or erhanedi ic acid) –
Dehydration.
- Reaction of cone. H2SO4 with an alcohol (other than methanol) – say ethyl alcohol or ethanol, c2H5OH
Dehydration.
- Reaction of cone. H2SO4 with hydrated copper (II) sulphate, CuS04. 5H2O (blue vitriol) – Dehydration.
Question 13.
State the observation seen when cone. H2SO4 is added to –
(1) sucrose
(2 ) hydrated copper (II) sulphate.
Answer:
- Cone. H2SO4 dehydrates sucrose to carbon, called sugar charcoal. This is m the form of a black spongy charged mass of carbon.
- Cone. H2S04 dehydrates blue crystals of copper (II) sulphate pentahydrate (blue vitriol) to copper sulphite, which is in the form of a white powder.
Question 14.
State how addition of –
- copper
- NaCl to hot cone. H2SO4 serves as a test for the latter.
Answer:
- Copper turnings when heated with cone. H2SO4 gives a colourless suffocating gas with a smell of burning sulphur(SO2). The gas turns orange coloured acidified potassium dichromate solution green. This can be used as a test for cone. H2SO4.
- Common salt (NaCl) when heated with cone. H2SO4 gives a colourless gas pungent smell only which (HCl) gives white fumes with NH3. This can be used as a test for cone. H2SO4.
Question 15.
Give two tests for dilute sulphuric acid with balanced equations. State why
(1) BaCl2
(2) Pb(NO3)2 are used for the above tests.
Answer:
- Barrium chloride solution on treating with dilute sulphuric acid forms white ppt. of barium sulphate, which is insoluble in all acids
BaCl2 + H2SO4 (dil) → 2HCl + BaS04 ↓ - Lead nitrate on treating with dilute sulphuric acid forms white ppt of lead sulphate, which is insoluble in all acids.
Pb (NO3)2 + H2SO4 (dil) → 2HNO3 + PbSO4 ↓
Question 16.
Give a test to distinguish dilute sulphuric acid from dilute HCl and dilute HNO3.
Answer:
Test to distinguish dil H2SO4 from dil HCl and HNO3- BaCl2 soln. when added to dil. H2S04 gives a white ppt. of BaSO4, but with dil. HCl and dil. HNO3, no white ppt. is produced – since BaCl2 and Ba(NO3)2 are soluble in dil. H2SO4.
Question 17.
State three different chemical compounds other than acids manufactured industrially from sulphuric acid.
Answer:
- Barium sulphate
- Lead sulphate
- Sodium sulphate
UNIT TEST PAPER 7D — H2SO4
Question 1.
Select the correct answer from the choice in brackets.
- The oxidised product obtained when sulphur reacts with cone. H2S04. (H2S/SO2/H2SO3).
Ans. SO2
- The dehydrated product obtained when cane sugar reacts with cone. H2S04. (CO / C / CO2)
Ans. C
- The type of salt formed when excess of caustic soda reacts with sulphuric acid, (acid salt / normal salt)
Ans. Normal salt
- The reduced product obtained when hydrogen sulphide reacts with cone. H2S04. (SO, / S / H.O)
Ans. SO2
- The salt which reacts with dil. H2SO4 acid to give an insoluble ppt. (Cu (NO3)2 / Zn (NOs)2 / Pb (NO3)2
Ans. Pb(NO3)2
Question 2
- Give a balanced equation for the conversion ‘A’.
Ans. 4FeS2 + 11O2→ 2Fe2O3 + 8SO2 - The gaseous mixture of the product of conversion ‘A’ and air contains dust particles as an impurity. Name another impurity in the same mixture.
Ans: Arsenious oxide (As2O3) - Is the conversion ‘B’ an exothermic or an endothermic reaction. Would lowering the temperature favour or retard the forward reaction.
Ans. Conversion of SO2 into SO3 is an exothermic reaction. As such lowering of temperature will favour the forward reaction i.e. Formation of SO3. - If the product of conversion ‘B’ is an acid anhydride of H2S04, the anhydride of conversion ‘A’ is………..
Ans. Acidic. - State why water is added for the conversion ‘D’ and not for the conversion ‘C’
Ans. SO3 is not directly dissolved in water to give H2SO4. This is because the dissolution of SO3 in water is highly exothermic resulting in production of dense fog of sulphuric acid particles which do not condense easily.
Question 3.
Give balanced equations for the following reactions using sulphuric acid.
- Formation of a black mark on a piece of wood on addition of cone. H2SO4 to it.
Ans. (C6H10O5)n + H2S04 (cone.) → 6(C)n + 5(H2O)n - Oxidation of a foul smelling acidic gas, heavier than air and fairly soluble in H2O by cone. H2SO4.
Ans. H2S + H,SO4 (cone.) → S + 2H2O + SO2 - Formation of an acid salt from sulphuric acid and (a) an alkali (b) a sodium salt.
Ans.
(a) Formation of an acid salt from sulphuric acid and an alkali (KOH)
KOH + H2SO4 (dil.)→ KHSO4 + 2H2O
(b) Formation of an acid salt from sulphuric acid and a sodium salt (NaCl)
- Formation of a hydrocarbon from an organic compound
Ans:
- Formation of sulphur dioxide using a metal below hydrogen in the activity series.
Ans: Cu + 2H2SO4 (cone.) → CuSO4 + 2H2O +SO2
Question 4.
Match the conversions in column ‘X’ using sulphuric acid, with the type of chemical property of sulphuric acid A to E it represents in column ‘Y’
Answer:
Question 5.
Select the correct substance from the substances A to J which react with the sulphuric acid to give the product 1 to 10. [State whether the acid used in each case is dilute or concentrated].
A : Iron
B : Sodium carbonate
C : Sodium chloride
D : Formic acid
E : Sodium nitrate
F : Sodium sulphite
G : Ethyl alcohol
H : Sodium sulphide
I : Sodium hydroxide (excess)
J : Hydrogen sulphide
- Product – ulphur dioxide
- Product- Sulphur
- Product-Hydrogen
- Product-Hydrochloric acid
- Product-Sodium sulphate
- Product-Carbon dioxide
- Product-Carbon monoxide
- Product-Nitric acid
- Product—Hydrogen sulphide
- Prorfwcf-Ethene
Answer:
- F: Sodium sulphite
- H: Sodium sulphide
- A: Iron
- C: Sodium chloride
- I: Sodium hydroxide
- E: Sodium Carbonate
- D: Formic acid
- E: Sodium nitrate
- H: Sodium sulphide
- G: Ithyl alcohol
Question 6.
Give reasons for the following:
- Sulphuric acid forms two types of salts with an alkali.
Ans. Sulphuric acid forms two types of . disc, viz., sulphates and bisulphates (or hydrogen sulphates) with alkales because it is a dibasic :acid, i.e. one molecule of H2SO4 on dissociation gives two H+ ions.
- Cone, sulphuric acid is used as a laLoraioiy reagent in the preparation of iodine from hydrogen iodide.
Ans. Cone. H2SO4. oxidises HI to iodine.
- Barium chloride solution can be used to distinguish between dil. H2SO4 and dil HNO3.
Ans. BaCl2 soln. when added to dil. H2SO4 gives a white ppt. of BaSO4, but with dil. HNO3, n: white ppt. is produced since BaCl2 and Ba(NO3), are soluble in dil. H2SO4. - The gaseous product obtained differs when zinc reacts with dilute and with cone. H2SO4
Ans. The metal reacts differently with dilute acid and concentrated acid. That is the character of metal. With dilute sulphuric acid, zinc gives hydrogen.
Zn + H2SO4 (dil.) → ZnSO4 + H2
The same metals will react differently with concentrated sulphuric acid to give sulphur dioxide gas.
- Ethanol can be converted to ethene using cone, sulphuric acid.
Ans. Ethanol cart be converted into ethene by heating it with cone. H2SO4 because cone. H2SO4 is a strong dehydrating agent.
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