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SOME BASIC CONCEPTS OF CHEMISTRY-3
1)The weight of calcium carbonate required to produce carbondioxide that is sufficient for conversion
of one 0.1 mole sodium carbonate to sodium bicarbonate is
1) 1gm
2) 10gm
3) 5gm
4)100gm
2.What weight of Oxygen required for the complete combustion of 2.4 grams of Mg?
1) 3.2gm
2) 32gm
3) 1.6gm
4)16gm
3. How many grams of 80% pure marble stone on calcination can give 14 grams of quick lime?
1) 25gm
2) 20gm
3) 31.25gm
4)11.2gm
4.. The percent weight loss suffered by sodium bicarbonate on strong heating is
1) 36.9%
2) 63.1%
3) 25.68%
4)73.8%
5.. 10 grams of a hydrated sodium carbonate, Na2CO3.xH2O, on strong heating suffers a weight loss
6.29 grams. The value of X is
1) 1
2) 6
3) 8
4)10
6.When 50 grams of sulphur was burnt in air, 4% of the impure residue is left over. Calculate the
volume of air required at STP containing 21% of oxygen by volume.
1)80lit
2)120lit
3)160lit
4)200lit
7.What is the weight of calcium carbonate required for the production of 1 L of carbon dioxide at
270C and 750mm, by the action of dilute hydrochloric acid?
1) 2.45gm
2) 4.009gm
3) 8 .018gm
4)40.09gm
8.The number of grams of NaOH that completely neutralises 4.9g of phosphoric acid is
1) 120
2) 24
3) 36
4) 6
9. The volume of oxygen required for the complete combussion of 10lit of methane under the same
conditions is
1) 20lit
2) 40lit
3) 10lit
4) 4lit
10. When 20ml of methane and 20ml of oxygen are exploded together and the reaction mixture is
cooled to laboratory temperature. The resulting volume of the mixture is
1) 40 ml
2) 20 ml
3) 30 ml
4) 10 ml
11. The volume of CO2 obtained by the complete decomposition of one mole of NaHCO3 at STP is
1) 22.4 L
2) 11.2 L
3) 44.8 L
4) 4.48 L
12. How many litres of CO2 at STP will be formed when 500 ml of 0.1M H2SO4 reacts with excess of
Na2CO3?
1) 1.12
2) 11.2
3) 0.224
4) 0.112
13 . 1g of Mg is burnt in a vessel containing 0.5g of oxygen. The reactant remaining unreacted is
1) 0.25g of Mg
2) 0.1g of Mg
3) 0.1g of O2
4) 0.75g of Mg
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14. A gas mixture contains acetylene and carbondioxide. 20 lit of this mixture requires 20 lit of
oxygen under the same conditions for complete combustion. The percentage by volume of
carbondioxide. in the mixture is
1) 50%
2) 40%
3) 60%
4) 75%
15. Acetylene can be prepared from calcium carbonate by a series of reactions. The mass of calcium
carbonate required to prepare 1mole of acetylene is
1) 200g
2) 100g
3) 250g
4) 320g
16. Sodium carbonate of 92% purity is used in the reation Na2CO3 + CaCl2→ CaCO3 + 2NaCl. The
number of grams of Na2CO3 required to yield 1gm of CaCO3
1) 8.5g
2) 10.5g
3) 11.52g
4) 1.152g
17. Benzene burns in oxygen according to the equation 2C6H6(l) + 15O2(g) →12CO2(g) + 6H2O(l).
How many litres of oxygen are required at STP for the complete combustion of 39g of liquid
benzene?
1) 11.2
2) 22.4
3) 42
4) 84
18. The mass of 80% pure H2SO4 required to completely neutralise 60g of NaOH is nearly
1) 92g
2) 58.8g
3) 73.5g
4) 98g
19. 6 grams of magnesite mineral on heating liberated carbondioxide which measures 1.12 L at
STP.The percentage purity of mineral is
1) 30%
2) 50%
3) 70%
4) 90%
20. The volume of hydrogen at STP required to reduce 7.95 grams of cupric oxide to metal is
1) 1.12lit
2) 11.2lit
3) 0.224lit
4) 2.24lit
21. What is the volume of ammonia obtained starting from 2 L of nitrogen, if the conversion is only
6% efficient in the given conditions?
1) 0.12lit
2) 1.2lit
3) 0.24 lit
4) 2. 4lit
22. A mixture of 26ml of H2 and 24ml of Cl2 is exploded. After reaction the tube contains unreact
1) 2ml of H2
2) 4ml of H2
3) 2ml of Cl2
4) 1ml of Cl2
23. To get 5.6 lit of CO2 at STP weight of CaCO3 to be decomposed is
1) 100g
2) 50g
3) 25g
4) 75g
24. The volume of chlorine that can completely react with 100mg of hydrogen at STP is
1) 1120 ml
2) 1.12 ml
3) 11.2 ml
4) 112 ml
25. 200 mL of pure and dry oxygen is subjected to silent electrical discharge. The volume of ozonised
oxygen obtained is 190 mL. What is the percent conversion of oxygen to ozone?
1) 3%
2) 5%
3) 10%
4) 15%
26. The volume of chlorine required for the complete reaction of 10 litres of H2S at STP is
1) 22.4 L
27.
2) 5 lit
3) 10 lit
4) 2.5 lit
6.54g of Zn is dissolved in excess of H2SO4. The weight of ZnSO4 formed and the volume of H2
gas liberated at STP are ( molar mass of Zn=65.4 g)
1) 161.4g, 2.24 L
2) 16.14g, 2.24 L
3) 16.14g, 22.4 L
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4) 16.14g, 11.2 L
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28. 10g of a mixture of CaCO3 and Na2CO3 on ignition suffered a loss in weight of 2.2g. The mass
ratio of CaCO3 and Na2CO3 is
1) 1:1
2) 1:1.4
3) 1.4:1
4) 1.75:1
29. A mixture of MgO and Mg weighing 10g is treated with excess of dil HCl. Then 2.24 lit of H2 gas
was liberated under STP conditions. The mass of MgO present in the sample is
1) 2.4g
2) 7.6g
3) 8g
4) 2g
30. One litre of a mixture of CO and CO2 is passed over red hot coke when the volume increased to
1.6 L under the same conditions of temperature and pressure. The volume of CO in the original
mixture is
1) 400 ml
2) 600 ml
3) 500 ml
4) 800 ml
KEY:
1)2
2) 3
3) 3
4) 1
11) 2 12) 1
13) 1
14) 3
21)3
23) 3
22) 1
5)4
24) 1
6) 3
7)2
15)2
16) 4
25)4
26) 3
8) 4
17) 4
27) 2
9) 3
18) 1
10) 2
19)3
28) 1
29)2
20)4
30)1
SOLUTIONS
1) The balanced equation for conversion of sodium carbonate to sodium bicarbonate is
Na2CO3 + H2O +
CO2
→
2NaHCO3
1 mole of Na2CO3 requires 1 mole i.e 44gm of CO2
Carbondioxide required for 0.1 mole of Na2CO3 = 4.4 grams
The balanced equation for the decomposition of calcium carbonate is
CaCO3
→
CaO + CO2
1 mole i.e 44gm of CO2 given by 1 mole i.e 100 grams of CaCO3
The weight of calcium carbonate to produce the required 4.4 grms carbondioxide
=(4.4X100/44)=10gms
2. Solution:
The balanced equation for the combustion of magnesium metal is
2Mg
+ O2
→
2MgO
2 moles of Mg = 1 mole of O2
2X24gm of Mg = 32gm of O2
2.4 grams of Mg= (2.4X32/2X24)= 1.6grams of O2
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3. Solution: CaCO3 → CaO + CO2
1 mole of CaO = 1 mole of CaCO3
56 grams of CaO given by 100 grams of CaCO3
14 grams of CaO given by ( 14/56)100=25 grams of CaCO3
As purity is 80%
i.e
80 grams of calcium carbonate is present in 100 grams of marble stone
25 grams of calcium carbonate is present in (25X100 /80)=31.25grams of marble stone
4.Solution: On strong heating sodium bicarbonate decomposes and loses carbondioxide and water.
2NaHCO3 →Na2CO3 + H2O + CO2
2 moles of NaHCO3 →1 mole of CO2 + 1 mole of H2O
(2 × 84) grams of NaHCO3 shows (44 + 18)=62grams weight loss
100 grams of NaHCO3 shows ( 100X62/2x84)= 36.9% weight loss.
5Solution: Hydrated sodium carbonate loses water of hydration on strong heating.
Na2CO3.xH2O → Na2CO3 + xH2O
1 mole of Na2CO3 → x moles of H2O
i.e 106 grams of Na2CO3 combines with 18x grams of water
in the given 10 gm , 3.71 grams of Na2CO3 combines with 6.29 grams of water
106 grams of Na2CO3 combines with 100X6.29/3.71)=180grams of water
18 x = = 180 therefore x=10
6 Sol :Weight of sample of sulphur taken = 50 g
Percentage of impurity = 4%
Weight of impurity in sample = ( 50X/100/4) = 2 grams
Weight of sulphur in sample = 50 – 2 = 48 grams
Combustion of sulphur gives sulphurdioxide
S + O2 → SO2
1 mole of S →1 mole of O2
32 grams of S = 22.4 L of O2 at STP
48 grams of S = ?
Volume of oxygen required at STP =(48/32) × 22.4 = 33.6 L
21 L of O2 is present in 100 L of air
33.6 L of O2 present in (33.6/21)X100== 160 lit of air.
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7. Sol :
Given conditions
P1 = 750 mm
P2 = 760 mm
T1 = 300K
T2 = 273K
V1 = 1 Lit
V2 = ?
STP conditions
From the equation of state, the unknown volume is obtained
P1V1 T2 750 × 1× 273
×
V2 = T1 P2 = 300 × 760 = 0.898 L
Calcium carbonate on reaction with hydrochloric acid gives carbondioxide.
CaCO3 + 2HCl →CaCl2 + H2O + CO2
1 mole of CO2 given by 1 mole of CaCO3
22.4 L of CO2 at STP = 100 grams of CaCO3
0.898 L of CO2 at STP = ?
100 × 0.898
22.4
= 4.009 grams
The weight of calcium carbonate required =
3NaOH + H3PO4 → Na3PO4 + 3H2O
8) Solution:
3molesi.e 3X40gm of NaOH neutralize 1mole i.e 98 gm of H3PO4
Wt of NaOH that neutrlise 4.9gm of H3PO4 =(4.9X3X40/98)=6gm
CH4 + 2O2→ CO2+ 2H2O
9. solution:
1mol Methane requires 2mol of Oxygen
i.e 1 lit of Methane requires 2lit of Oxygen
10lit of Methane requires =10X2=20lit of Oxygen.
10.
Solution:
CH4 + 2O2→ CO2+ 2H2O
1mol Methane reacts with 2mol of Oxygen to give 1mol of CO2
i.e 1 ml of Methane reacts with 2ml of Oxygen to give 1ml of CO2
10m of Methane reacts with 20ml of Oxygen to give 10ml of CO2.
Vol. of methane left=10ml, vol of CO2 formed=10ml
Total vol =20ml
11. Soltion: 2NaHCO3 →Na2CO3 + H2O + CO2
2 moles of NaHCO3 →1 mole i.e 22.4 lit of CO2 STp
1 mole of NaHCO3 gives 22.4/2 =11.2 lit of CO2 STp
12. Solution: moles of H2SO4 in solution = MX Vin lit =0.1 X500/1000 )=0.05
Na2CO3 + H2SO4→
Na2SO4 + H2O + CO2
1mole of sulphuric acid gives 1mole i.e 22.4 lit of CO2
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0.05 moles of sulphuric acid gives 0.05X22.4 = 1.12 lit of CO2
13 . Solution:
2Mg + O2 →
2 MgO
2moles i.e 2X24 gm of Mg reacts with 1mole i.e 32 gm of Oxygen.
Wt of Mg that reacts with 0.5gm of Oxygen =( 0.5X2X24/32)=0.75gm
Wt of Mg left +1- 0.75 = 0.25gm
14. Solution:
CO2 is not combustable. If vol of C2H2 =x then vol. of CO2=20-x
2 C2H2 +5O2→4 CO2 +2H2O
2 vol.of acetylene reacts with 5 vol.of oxygen
2lit of acetylene reacts with 5 lit .of oxygen
? of acetylene reacts with 20 lit .of oxygen
Vol. of acetylene =(20X2/5)=8lit
Vol.of CO2 in the mixture= 20-8 =12lit
Vol% of Co2 = ( 12X100/20)=60%
15. Solution:
CaCO3+3C→ CaC2+3CO
CaC2+2H2O→ C2H2+Ca(OH)2 , as per equations 1mole i.e 100gm of CaCO3 gives 1mole of
C2H2
16:Solution
100gm of CaCO3 given by 106 gm of Na2CO3
1gmof CaCO3 given by (106/100)=1.06gm of Na2CO3
Wt of 92% Na2CO3 required =(1.06 X100/92)=1.152gm .
17. Solution: 2C6H6(l) + 15O2(g) → 12CO2(g) + 6H2O(l).
2moles i.e 2X78=156 gmof Benzene requires 15X22.4 lit of oxygen at STP
39 gm of Benzene requires (39X15X22.4/156)= 84 lit of oxygen at STP
18. Solution:
2NaOH + H2SO4 → Na2SO4 + 2H2O
2moles i.e 2X40=80gm NaOH is given by 98gm of H2SO4
60gm NaOH is given by (60X98/80)= 73.5gm of H2SO4
Wt.of 80% acid required= (73.5X100/80)= 91.875gm
19. Solution: magnesite is magnesium carbonate. On heating it liberates carbondioxide.
MgCO3 →MgO + CO2
1 mole of CO2 is given by 1 mole of MgCO3
22.4 L of CO2 at STP is given by 84 grams of MgCO3
1.12 L of CO2 at STP = ?
The weight of pure MgCO3 to be decomposed =(1.12 × 84/22.4) = 4.2 grams
6 grams mineral has 4.2 grams MgCO3
Percent purity = × 100 = 70%
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20. Solution: Cupric oxide is reduced to copper on treating with hydrogen
CuO + H2 →Cu + H2O
1 mole of CuO requires 1 mole of H2
79.5 grams of CuO requires 22.4 L of H2 at STP
7.95 grams of CuO requires ( 7.95 × 22.4/79.5) = 2.24 L of hydrogen required at
STP
21. Solution: Conversion of nitrogen to ammonia is given as
N2 + 3H2 → 2NH3
1 mole of N2 = 2 moles of NH3
1 volume of N2 = 2 volumes of NH3
Volume of ammonia with 100% efficient conversion of 2 L nitrogen = 4 L
Volume of ammonia with 6% efficient conversion of the reaction = (6X4/100) = 0.24 L
22. Solution:
H2 + Cl2→
2HCl
1 vol.of H2 reacts with 1 vol. of Cl2
1 ml.of H2 reacts with 1 ml. of Cl2
?
ml.of H2 reacts with 24 ml. of Cl2
vol. of H2 reacts =24ml.
vol. of H2 left =26-24= 2ml.
23.
Solution: CaCO3 → CaO+ CO2
1mole i.e 22.4 lit of CO2 is given by 1mole i.e 100 gm CaCO3
5.6 lit of CO2 is given by (5.6X100 /22.4)=25 gm CaCO3
24. Solution:
H2 + Cl2→
2HCl
1 mol.of H2 reacts with 1 mol. of Cl2
2 gm i.e 2000mg of H2 reacts with 22400 ml. of Cl2
100mg of H2 reacts with (100X22400/2000) =1120ml. of Cl2
25.Solution:
3O2
→
3 vol
200 ml
(200-3x) mL
2O3
2 vol
0 mL
2x mL
Ozonised oxygen = O2 remaining + O3 obtained = 190 mL = (200 – 3x + 2x) ml
Solving, x = 10 mL
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volume of O2 remaining = 200 – 30 = 170 mL
volume of O2 consumed = 30 mL
If 200 mL O2 is taken = 30 mL is converted
If 100 mL O2 is taken = ?
The percentage conversion of oxygen = × 30 = 15%
26.Solution:
Cl2 + H2S→ 2HCl + S
1 vol.of chlorine reacts with 1 vol.of H2S
1 lit .of chlorine reacts with 1 lit .of H2S
10lit .of chlorine reacts with 10lit .of H2S
27.Solution : Zn +H2SO4
→ H2 +ZnSO4
1mole i.e 65.4 gm of Zn gives 1mole i.e 161.4 gmof ZnSO4 and 22. 4 lit H2 at STP
6.54 gm of Zn gives 16.14 gmof ZnSO4 and 2 .24 lit H2 at STP
28. Solution: Na2CO3 is thermally stable and does not decompose on ignition but CaCO3 decompse
to give CO2.
→
CaCO3
CaO + CO2
1mole i.e 100gm of CaCO3 gives 44 gm of CO2.
Wt.of CaCO3 giving 2.2 gm of CO2 =( 2.2X100/44)=5gm
Wt.of CaCO3 =5gm, Wt.of Na2CO3=10-5=5gm, There foe ratio of wts=1:1
29.Solution: Among Mg and MgO ,only Mg gives Hydrogen gas with acid .
Mg + 2HCl→ MgCl2+H2
1 mole i.e 24gm of ‘Mg’ gives 22.4 lit of H2 at STP.
Wt .of Mg giving 2.24 lit of H2 at STP=( 2.24X24/22.4)=2.4gm
Wt .of Mg=2.4gm and wt of MgO in the mixture= 10-2.4= 7.6gm.
30.Solution: Among CO and CO2 , only CO2 reacts with coke.
Vol .of CO2 =X lit then vol. of CO = 1-X
CO2 +C → 2CO
1 vol of CO2 gives 2 vol .of CO
X lit of CO2 gives 2X lit .of CO.
Final vol of mixture= (1-X)+2X= 1+X=1.6 lit i.e X=0.6lit
Therefore volume of ‘CO’ in the mixture = 1-X= 1-0.6= 0.4lit i.e 400ml
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