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Chemistry - Mole Concept and
Stoichiometry (A) (Solved)
Num.1.
2H2O → 2H2 (g) + O2 (g).
If a given experiment results in 2500 cm3 of H2 being produced, what volume of O2 is liberated at
the same time under the same conditions of temperature and pressure.
Solution : When 2 vol. of hydrogen is produced the 1 vol. of oxygen is liberated.
Therefore, When 2500 cm3 of hydrogen is produced then 1/2×2500 cm 3 of oxygen is liberated.
Hence, volume of oxygen liberated = 1250 cm 3.
Num.2.
4NH3 + 5O2 → 4NO + 6H2O.
If 27 litres of reactants are consumed, what volume of nitrogen monoxide (Nitric oxide) is
produced at the same temperature and pressure.
Solution : Volume of reactants = 4 vol. of ammonia + 5 vol. of oxygen = 9 vol.
9 vol. of reactants produces 4 vol. of Nitric oxide
Therefore, 27 vol. of reactants will produce 4/9×27 lit. = 12 litres of Nitric oxide.
Num.3.
4N2O + CH4 → CO2 + 2H2O + 4N2.
If all volumes are measured at the same temperature and pressure. Calculate the volume of N 2O
required to give 150 cm3of steam.
Solution : 2 vol. of steam is produced by 4 vol. of N2O
1 vol. will be produced by 4/2 vol. of N2O
150 cm3 of steam will be produced by 4/2×150 = 300 cm3 of N2O.
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Num.4. What volume of oxygen would be required for the complete combustion of 100 litres of
ethane according to the following equation.
2C2H6 + 7O2 → 4CO2 + 6H2O.
Solution : 2 vol. of ethane requires 7 vol. of oxygen
1 vol. of ethane will require 7/2 vol. of oxygen
100 litres of ethane will require 7/2 × 100 = 350 litres of oxygen.
Num.5. What vol. of O2 is required to burn completely a mixture of 22.4 dm3 of CH4 and 11.2 dm3 of
H2. The reaction are :
CH4 + 2O2 → CO2 + 2H2O ;
2H2 + O2 → 2H2O.
Solution : CH4 + 2O2 → CO2 + 2H2O
1 vol. 2 vol. 1 vol.
1 vol. of CH4 requires 2 vol. of O2
22.4 dm3 of CH4 will require 2 × 22.4 = 44.8 dm3 -------- (i)
2H2 + O2 → 2H2O
2 vol. 1 vol. 2 vol.
2 vol. of H2 requires 1 vol. of O2
1 vol. of H2 will require ½ vol. of O2
11.2 dm3 of H2 will require 1/2×11.2 = 5.6 dm3 ---------- (ii)
Total vol. = 44.8 + 5.6 = 50.4 dm3.
Num.20. Find the total percentage of oxygen in magnesium nitrate crystal :
Mg(NO3)2.6H2O [H = 1, N = 14, O = 16, Mg = 24].
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Solution : gram molecular mass of Mg(NO3)2.6H2O = 24 + (14 + 48) × 2 + 6 × 18 = 256.
Mass of oxygen in magnesium nitrate = 192 g
% of oxygen = (192 / 256) × 100 = 75 %.
Num.21. What is the mass of nitrogen in 1000 kg of urea [CO(NH2)2]. [Answer to the nearest kg.]
Solution : Molecular weight of urea [CO(NH2)2] = 12 + 16 + (14 + 2) × 2 = 60.
60 kg of urea contains 28 kg of nitrogen
1000 kg of urea will contain (28/60) × 1000 = 467kg of nitrogen .
Num.22. Calculate the percentage of boron [B] in borax Na2B4O7.10H2O [H = 1, B = 11, O = 16, Na =
23].
Solution :
Num.23. If a crop of wheat removes 20 kg of nitrogen per hectare of soil, what mass of the fertilizer
calcium nitrate Ca(NO3)2 would be required to replace the nitrogen in a 10 hectare field? [N = 14; O = 16;
Ca = 40] (Give your answer to the nearest kg).
Solution : Click Here
Num.6. 560 ml of carbon monoxide is mixed with 500 ml of oxygen and ignited. The chemical
equation for the reaction is as follows:
2CO + O2 → 2CO2.
Calculate the volume of oxygen used and carbon dioxide formed in the above reaction.
Solution : 2CO + O2 → 2CO2
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2 vol. 1 vol. 2 vol.
To burn 2 vol. of CO, 1 vol. of O2 is required and 2 vol. of CO2 is produced
To burn 560 ml of CO, 560/2 = 280 ml of O2 will be required and 560 ml of CO2 will be produced.
Hence, vol. of O2 required = 280 ml, vol. of CO2produced = 560 ml.
Num.7. Each of two flasks contains 2.0 g of gas at the same temperature and pressure. One flask
contains oxygen and the other hydrogen. (a) Which sample contains the greater number of
molecules. (b) If the H2 sample contains N molecules, how many are in the O2 sample. [H = 1; O =
16].
Solution : (a) 2 g of H2will have greater number of molecules.
2 g of H2 = 1 mole = 6.023 × 1023 molecules.
2 g of O2 = 2/32 mole = 1/16 mole = (6.023 × 1023)/16 molecules.
[1 mole of O2 = 32 g => 2 g of O2 = 2/32 mole]
Hence the answer.
(b) No of molecules of oxygen = 2/32 ×N = N/16.
Num.8. 112 cm3 of gaseous fluoride of phosphorus has a mass 0.63 g. Calculate it’s relative
molecular mass. If the molecule of the fluoride contains only one atom of phosphorus, then
determine the formula of the phosphorus fluoride. [F = 19; P = 31].
Solution : Weight of 112 cm3 of gaseous fluoride is 0.63g
Weight of 22400 cm3 = 0.63/112 × 22400 g = 126 g.
Let PFn = 126
Or, 31 + 19n = 126
Or, 19n = 126 – 31 = 95
Or, n = 95/19 = 5
Hence, formula of Phosphorus fluoride is PF5.
Num.9. When heated, potassium permanganate decomposes according to the following equation:
2KMnO4 → K2MnO4 + MnO2 + O2.
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Given that the molecular mass of potassium permanganate is 158, what volume of oxygen
(measured at room temperature) would be obtained by the complete decomposition of 15.8 g of
potassium permanganate? (Molar volume at room temperature is 24 litres.)
Solution : Molar mass of KMnO4 = 158
Molar vol. of O2 at room temp. = 24 litre
158 g of KMnO4 at room temp. yields 24 litres of O2
15.8 g of KMnO4 at room temp. will yield = 24/(2 × 158) × 15.8
= 1.2 litres of O2.
Num.10. A flask contain 3.2 g of Sulphur dioxide. Calculate the following:
(i) The moles of Sulphur dioxide present in the flask.
(ii) The number of molecules of Sulphur dioxide present in the flask.
(iii) The volume occupied by 3.2 g of Sulphur dioxide at S. T. P. (S = 32, O = 16).
Solution : (i) Moles = Weight of substance in grams / Molecular weight
Hence, mole of SO2 = 3.2 / 64 = 0.05.
(ii) Number of molecules = Moles × 6.023 × 1023
Hence, number of molecules of SO2 = 0.05 × 6.023 × 1023 = 0.302 × 1023.
(iii) 64 gms of SO2 occupies a vol. of 22.4 lit.
3.2 gms of SO2 will occupy (3.2 × 22.4) / 64 = 1.12 litres.
Num.11. The volume of gases A, B, C and D are in the ratio, 1 : 2 : 2 : 4 under the same conditions
of temperature and pressure:
(i) Which sample of gas contains the maximum number of molecules?
(ii) If the temperature and the pressure of gas A are kept constant, then what will happen to the
volume of A when the number of molecules is doubled?
(iii) If this ratio of gas volumes refers to the reactants and products of a reaction, which gas law is
being observed?
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(iv) If the volume of A is actually 5.6 dm3 at S. T. P. calculate the number of molecules in the actual
volume of D, at S. T. P. (Avogadro’s Number is 6 × 1023).
(v) Using your answer from (iv) state the mass of D if the gas is Di-nitrogen dioxide (N2O). (N = 14;
O = 16).
Solution : (i) The sample D. (ii) Volume of A will get doubled. (iii) Gay Lussac’s Law. (iv)
Gases
Volumes
A
1
D
:
5.6 dm3
4
4 × 5.6 dm3 at S. T.
P.
= 22.4 dm3(Molar
volume)
= 6 × 1023molecules.
(v) 6 × 1023 molecules is Avogadro’s number of molecules contained in 1 gram mole of the substance. If gas d is N 2O
then ,
1 gram mole of N2O = 2 × 14 + 16 = 44g.
Num.12. (i) Calculate the number of moles and the number of molecules present in 1.4 g of
ethylene gas. What is the volume occupied by the same amount of ethylene?
(ii) What is the vapour density of ethylene? (Avogadro’s Number = 6 × 1023; Atomic weight of C =
12, H = 1;
Molar volume = 22.4 litres at S. T. P.)
Solution :
(i) Molecular formula of ethylene = CH2 = CH2
Molecular weight of ethylene = 12 + 2 + 12 + 2 = 28 g
Number of moles = Given weight / Molecular weight
= 1.4 / 28 = 0.05 moles.
Number of molecules in 1 mole = 6 × 1023
Therefore, number of molecules in 0.05 mole = 6 × 1023 × 0.05 = 0.3 × 1023
= 3 × 1022 molecules.
Volume occupied by 1 mole of ethylene = 22.4 liters
Therefore, vol. occupied by 0.05 mole of ethylene = 22.4 × 0.05 = 1.12 litres.
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(ii) Vapour density, V.D. = Molecular weight /2 = 28/2 = 14.
Num.13. A sample of ammonium nitrate when heated yields 8.96 litres of steam (measured at
S.T.P.).
NH4NO3 → N2O + 2H2O
(i) What volume of dinitrogen oxide is produced at the same time as 8.96 litres of steam?
(ii) What mass of ammonium nitrate should be heated to produce 8.96 litres of steam? (Relative
molecular mass of ammonium nitrate is 80).
Solution : NH4NO3 → N2O + 2H2O
→1 vol. (22.4 L) + 2 vol.(44.8L)
(i) When 44.8 L of steam is produced then 22.4 L of N2O is produced at stp.
When 8.96 L of steam is produced then (22.4 × 8.96)/44.8 L = 4.48 L of N2O at stp.
(ii) 44.8 L steam is liberated by 80 g NH4NO3
8.96 L steam will be liberated by (80 × 8.96)/44.8 = 16 g NH4NO3.
Num.14. Under the same conditions of temperature and pressure, you collect 2 litres of Carbon
dioxide, 3 litres of Chlorine, 5 litres of Hydrogen, 4 litres of Nitrogen and 1 litre of Sulphur dioxide.
In which gas sample will there be:
(i) the greatest number of molecules?
(ii) the least number of molecules?
Justify your answer.
Solution : (i) Hydrogen.
(ii) Sulphur dioxide.
Justification : According to Avogadro’s Law, under the similar condition of temperature and pressure, equal volume
of all gases contain equal number of molecules. Therefore volume of Hydrogen is greater and so, it has greater
number of molecules.
Num.15. The gases Chlorine, Nitrogen, Ammonia and Sulphur dioxide are collected under the
same condition of temperature and pressure. Copy the following table which gives the volume of
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gases collected and the number of molecules (X) in 20 litres of Nitrogen. You are required to
complete the table giving the number of molecules in the other gases in terms of X :
Gas
Volume (litres)
Chlorine
10
Nitrogen
20
Ammonia
20
Sulphur dioxide
5
Gas
Volume (litres)
Number of
molecules
Chlorine
10
X/2
Nitrogen
20
X
Ammonia
20
X
Sulphur dioxide
5
X/4
Number of
molecules
X
Solution :
Num.16. Samples of the gases O2, N2, CO2 and CO under the same conditions of temperature and
pressure contain the same number of molecules represented by X. The molecules of Oxygen (O 2)
occupy V litres and have a mass of 8 gms. Under the same conditions of temperature and
pressure:
(i) What is the volume occupied by:
(1) X molecules of N2
(2) 3X molecules of CO?
(ii) What is the mass of CO2 in grams?
(iii) In answering the above questions, whose law has been used?
(C = 12, N = 14, O = 16).
Solution : (i) (1) V litres (2) 3V litres
(ii) Number of mol of oxygen = Mass of O2/Molar mass of O2 = 8 g/ 32 g mol-1
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= 0.25 mol.
As the number of molecules of CO2 is equal to the number of moles of O2 = 0.25 mol.
Mass of CO2 in grams = 0.25 × 44 = 11 gms
(iii) Avogadro’s Law.
Num.17. When heated, potassium permanganate decomposes according to the following
equation :
2KMnO4 → K2MnO4 + MnO2 + O2
Some potassium permanganate was heated in a test tube. After collecting one litre of oxygen at
room temperature, it was found that the test-tube had undergone a loss in mass of 1.32 g. If one
litre of hydrogen under the same condition of temperature and pressure has a mass of 0.0825 g,
calculate the relative molecular mass of oxygen.
Solution : Mass of 1 lit. of oxygen = 1.32 g
Mass of 1 lit. of hydrogen = 0.0825 g
Vapour density of oxygen = mass of 1 lit. of oxygen / mass of 1 lit. of hydrogen
= 1.32 /0.0825 = 16 g.
Relative molecular mass of oxygen = 2 × vapour density = 2 × 16 = 32.
Q.18. Name the term which defines the mass of a given volume of a gas compared to the mass of
an equal volume of hydrogen.
Solution : Vapour density.
Num.19. Find the relative molecular mass of a gas, 0.546 g of which occupies 360 cm3 at 87 ºC
and 380 mm Hg pressure. [1 litre of hydrogen at s.t.p. weigh 0.09 g].
Solution : P1 = 300 mm of Hg, V1 = 360 cm3 = 360 ml, T1 = 87 + 273 = 360 K.
P2 = 760 mm of Hg, V2 = x ml, T2 = 273 K
Using, P1 V1 /T1 = P2 V2 / T2 we get
(300 × 360) / 360 = (760 × x) / 273 => x = (273 × 300) / 760 = 17.7631 l
Mass of 17.7631 ml = 0.546 g
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=> mass of 1000 ml = (0.546/17.7631) × 1000 = 1.69 g
Vapour density = mass of 1 lit. of the gas / mass of 1 lit. of hydrogen
= 1.69/0.09 = 18.78
Molecular mass = 2 × 18.78 = 37.56.