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Stoichiometric Calculations
1. The weight of calcium carbonate required to produce carbon-dioxide that is
sufficient for conversion of one 0.1 mole sodium carbonate to sodium bicarbonate
1) 1gm
2) 10gm
co
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is
3) 5gm
4)100gm
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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
4) 11.2gm
The percent weight loss suffered by sodium bicarbonate on strong heating is
2) 63.1%
3) 25.68%
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1) 36.9%
5.
3) 31.25gm
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4.
2) 20gm
4)73.8%
10 grams of a hydrated sodium carbonate, Na2CO3.xH2O, on strong heating
1) 1
.s
a
suffers a weight loss of 6.29 grams. The value of X
2) 6
3) 8
is
4) 10
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6. When 50 grams of sulphur was burnt in air, 4% of the impure residue is left over.
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Calculate the volume of air required at STP containing 21% of oxygen by volume.
1) 80lit
7.
2) 120lit
3) 160lit
4) 200lit
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
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4) 40.09gm
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8. The number of grams of NaOH that completely neutralizes 4.9 g of phosphoric
acid is
1) 120
2) 24
3) 36
4) 6
9. The volume of oxygen required for the complete combustion of 10lit of methane
1) 20lit
2) 40lit
3) 10lit
co
m
under the same conditions is
4) 4lit
10. When 20ml of methane and 20ml of oxygen are exploded together and the
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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
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12. How many liters of CO2 at STP will be formed when 500 ml of 0.1M H2SO4
2) 11.2
.s
a
1) 1.12
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reacts with excess of Na2CO3?
3) 0.224
4) 0.112
13. 1g of Mg is burnt in a vessel containing 0.5g of oxygen. The reactant remaining
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unreacted is
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1) 0.25g of Mg
2) 0.1g of Mg
3) 0.1g of O2
4) 0.75g of Mg
14. A gas mixture contains acetylene and carbon dioxide. 20 lit of this mixture require
20 lit of oxygen under the same conditions for complete combustion. The
percentage by volume of carbon dioxide in the mixture is
1) 50%
2) 40%
3) 60%
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4) 75%
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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 reaction
to yield 1gm of CaCO3
1) 8.5g
2) 10.5g
3) 11.52g
4) 1.152g
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17. Benzene burns in oxygen according to the equation
co
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Na2CO3 + CaCl2→ CaCO3 + 2NaCl. The number of grams of Na2CO3 required
2C6H6(l) + 15O2(g) →12CO2(g) + 6H2O(l). How many liters 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
2) 58.8g
3) 73.5g
4) 98g
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1) 92g
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19. 6 grams of magnesite mineral on heating liberated carbon dioxide which
.s
a
measures 1.12 L at STP. The percentage purity of mineral is
1) 30%
2) 50%
3) 70%
4) 90%
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20. The volume of hydrogen at STP required to reduce 7.95 grams of cupric oxide to
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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) 0.24lit
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4) 2. 4lit
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22. A mixture of 26ml of H2 and 24ml of Cl2 is exploded. After reaction the tube
contains unreacted
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
2) 50g
3) 25g
4) 75g
co
m
1) 100g
24. The volume of chlorine that can completely react with 100mg of hydrogen at STP
1) 1120 ml
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is
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%
1) 22.4 L
2) 5 lit
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STP is
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26. The volume of chlorine required for the complete reaction of 10 liters of H2S
3) 10 lit
at
4) 2.5 lit
.s
a
27. 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)
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1) 161.4g, 2.24 L 2) 16.14g, 2.24 L
3) 16.14g, 22.4 L
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 CaCO 3 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
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1) 2.4g
2) 7.6g
3) 8g
4) 2g
30. One liter 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
2) 600 ml
3) 500 ml
4) 800 ml
co
m
1) 400 ml
1) 2
2) 3
3) 3
4) 1
11) 2
12) 1
13) 1
21) 3
22) 1
23) 3
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KEY
5)4
6) 3
7) 2
8) 4
9) 3
10) 2
14) 3 15) 2
16) 4
17) 4
18) 1
19) 3
20) 4
24) 1 25) 4
26) 3
27) 2
28) 1
29) 2
30) 1
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SOLUTIONS
1) The balanced equation for conversion of sodium carbonate to sodium bicarbonate is
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Na2CO3 + H2O +
CO2
→
2NaHCO3
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a
1 mole of Na2CO3 requires 1 mole i.e. 44gm of CO2
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Carbon dioxide required for 0.1 mole of Na2CO3 = 4.4 grams
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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 carbon dioxide
= (4.4X100/44) =10gms
2. Solution:
The balanced equation for the combustion of magnesium metal is
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2Mg
+ O2
→
2MgO
2 moles of Mg = 1 mole of O2
2X24gm of Mg = 32gm of O2
3. Solution: CaCO3 → CaO + CO2
1 mole of CaO = 1 mole of CaCO3
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56 grams of CaO given by 100 grams of CaCO3
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2.4 grams of Mg= (2.4X32/2X24) = 1.6grams of O2
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
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and water.
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4. Solution: On strong heating sodium bicarbonate decomposes and loses carbon dioxide
.s
a
2NaHCO3 →Na2CO3 + H2O + CO2
2 moles of NaHCO3 →1 mole of CO2 + 1 mole of H2O
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(2 × 84) grams of NaHCO3 shows (44 + 18) =62grams weight loss
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100 grams of NaHCO3 shows (100X62/2x84) = 36.9% weight loss.
5. Solution: Hydrated sodium carbonate loses water of hydration on strong heating.
Na2CO3.xH2O → Na2CO3 + xH2O
1 mole of Na2CO3 → x moles of H2O
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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
6 Sol: Weight of sample of sulphur taken = 50 g
Percentage of impurity = 4%
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Weight of impurity in sample = (50X/100/4) = 2 grams
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18 x = = 180 therefore x=10
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
48 grams of S =?
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32 grams of S = 22.4 L of O2 at STP
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Volume of oxygen required at STP = (48/32) × 22.4 = 33.6 L
.s
a
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
STP conditions
P2 = 760 mm
T1 = 300K
T2 = 273K
V1 = 1 Lit
V2 =?
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P1 = 750 mm
From the equation of state, the unknown volume is obtained
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V2
P1V1 T2

= T1 P2
=
750  1 273
300  760
= 0.898 L
Calcium carbonate on reaction with hydrochloric acid gives carbon dioxide.
CaCO3 + 2HCl →CaCl2 + H2O + CO2
co
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1 mole of CO2 given by 1 mole of CaCO3
22.4 L of CO2 at STP = 100 grams of CaCO3
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0.898 L of CO2 at STP =?
The weight of calcium carbonate required =
100  0.898
22.4
= 4.009 grams
8. Solution: 3NaOH + H3PO4 → Na3PO4 + 3H2O
3molesi.e 3X40gm of NaOH neutralize 1mole i.e. 98 gm of H3PO4
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Wt of NaOH that neutralize 4.9gm of H3PO4 =(4.9X3X40/98)=6gm
CH4 + 2O2→ CO2+ 2H2O
9. Solution:
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1mol Methane requires 2mol of Oxygen
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a
i.e 1 lit of Methane requires 2lit of Oxygen
10lit of Methane requires =10X2=20lit of Oxygen.
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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 CO 2
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
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11. Solution: 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
Na2CO3 + H2SO4→
Na2SO4 + H2O + CO2
co
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12. Solution: moles of H2SO4 in solution = MX Vin lit =0.1 X500/1000) =0.05
1mole of sulphuric acid gives 1mole i.e. 22.4 lit of CO2
13. Solution: 2Mg + O2 →
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0.05 moles of sulphuric acid gives 0.05X22.4 = 1.12 lit of CO2
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
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2 C2H2 +5O2→4 CO2 +2H2O
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2 vol. of acetylene reacts with 5 vol. of oxygen
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2lit of acetylene reacts with 5 lit .of oxygen
Of acetylene reacts with 20 lit .of oxygen
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Vol. of acetylene = (20X2/5) =8lit
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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
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16. Solution
100gm of CaCO3 given by 106 gm of Na2CO3
1gmof CaCO3 given by (106/100) =1.06gm of Na2CO3
17. Solution: 2C6H6 (l) + 15O2 (g) → 12CO2 (g) + 6H2O (l).
co
m
Wt of 92% Na2CO3 required = (1.06 X100/92) =1.152gm.
2moles i.e. 2X78=156 gm of Benzene requires 15X22.4 lit of oxygen at STP
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39 gm of Benzene requires (39X15X22.4/156) = 84 lit of oxygen at STP
2NaOH + H2SO4 → Na2SO4 + 2H2O
18. Solution:
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 carbon dioxide.
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MgCO3 →MgO + CO2
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1 mole of CO2 is given by 1 mole of MgCO3
.s
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22.4 L of CO2 at STP is given by 84 grams of MgCO3
1.12 L of CO2 at STP =?
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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%
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
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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
1 mole of N2 = 2 moles of NH3
1 volume of N2 = 2 volumes of NH3
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N2 + 3H2 → 2NH3
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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
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Vol. of H2 reacts =24ml.
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Vol. of H2 left =26-24= 2ml.
.s
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23. Solution: CaCO3 → CaO+ CO2
1mole i.e. 22.4 lit of CO2 is given by 1mole i.e. 100 gm CaCO3
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5.6 lit of CO2 is given by (5.6X100 /22.4) =25 gm CaCO3
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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
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→
3O2
2O3
3 vol
2 vol
200 ml
0 mL
(200-3x) mL
2x mL
Solving, x = 10 mL
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Volume of O2 remaining = 200 – 30 = 170 mL
co
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Ozonised oxygen = O2 remaining + O3 obtained = 190 mL = (200 – 3x + 2x) 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%
Cl2 + H2S→ 2HCl + S
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26. Solution:
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1 vol. of chlorine reacts with 1 vol. of H2S
1 lit .of chlorine reacts with 1 lit .of H2S
.s
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10lit .of chlorine reacts with 10lit .of H2S
→ H2 +ZnSO4
27. Solution: Zn +H2SO4
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1mole i.e. 65.4 gm of Zn gives 1mole i.e. 161.4 gm of ZnSO4 and 22. 4 lit H2 at STP
6.54 gm of Zn gives 16.14 gm of ZnSO4 and 2 .24 lit H2 at STP
28. Solution: Na2CO3 is thermally stable and does not decompose on ignition but CaCO 3
decompose to give CO2.
CaCO3 →
CaO + CO2
1mole i.e. 100gm of CaCO3 gives 44 gm of CO2.
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Wt. of CaCO3 giving 2.2 gm of CO2 = (2.2X100/44) =5gm
Wt. of CaCO3 =5gm, Wt. of Na2CO3=10-5=5gm. Therefore, ratio of wts=1:1
29. Solution: Among Mg and MgO, only Mg gives Hydrogen gas with acid.
1 mole i.e. 24gm of ‘Mg’ gives 22.4 lit of H2 at STP.
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Mg + 2HCl→ MgCl2+H2
Wt .of Mg giving 2.24 lit of H2 at STP= (2.24X24/22.4)=2.4gm
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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.
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Final vol. of mixture= (1-X)+2X= 1+X=1.6 lit i.e. X=0.6lit
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Therefore volume of ‘CO’ in the mixture = 1-X= 1-0.6= 0.4lit i.e. 400ml
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