WBBSE Class 10 Physical Science Solutions Chapter 3 Chemical Calculations

Detailed explanations in West Bengal Board Class 10 Physical Science Book Solutions Chapter 3 Chemical Calculations offer valuable context and analysis.

WBBSE Class 10 Physical Science Chapter 3 Question Answer – Chemical Calculations

Very Short Answer Type Questions

Question 1.
What is the molecular weight of ammonium phosphate ?
Answer:
149

Question 2.
How much CaO is obtained from 10 kg CaCO3 ?
Answer:
5.6 kg

Question 3.
How much CaCO3 will react with dil HCl to produce 22 g CO2 ?
Answer:
50 g

WBBSE Class 10 Physical Science Solutions Chapter 3 Chemical Calculations

Question 4.
How much hydrogen is required to produce 72 g water?
Answer:
8g

Question 5.
How much potassium chlorate is to be heated to produce as much oxygen as required to burn 6 g carbon completely?
Answer:
40.83 g

Question 6.
How much quantity of silver chloride (AgCl) can be obtained from 1.0 g silver nitrate?
Answer:
0.844 g

Question 7.
How many grams limestone are needed to get 48g CO2 ?
Answer:
109.09 g

Question 8.
How much hydrogen is produced when steam is passed over 28 kg red hot iron?
Answer:
1.33 kg

WBBSE Class 10 Physical Science Solutions Chapter 3 Chemical Calculations

Question 9.
What is mole number of a substance ?
Answer:
Mole number of a substance

WBBSE Class 10 Physical Science Solutions Chapter 3 Chemical Calculations 1

Question 10.
What is the percentage of water in blue vitriol ?
Answer:
36.07%.

Question 11.
2 Mg + O2 = 2 MgO; what type of chemical reaction is it?
Answer:
2 Mg + O2 = 2 MgO is direct combination type of chemical reaction.

Question 12.
HCl + NaOH = NaCl + H2 O; what type of chemical reaction is it?
Answer:
HCl + NaOH = NaCl + H2 O is the type of neutralisation reaction.

Question 13.
Zn + CuSO4 = ZnSO4 + Cu; what type of chemical reaction is it?
Answer:
Zn + CuSO4 = ZnSO4 + Cu is the type of substitution reaction.

WBBSE Class 10 Physical Science Solutions Chapter 3 Chemical Calculations

Question 14.
NH4 CNO (8) CO(NH2)2; what type of chemical reaction is it?
Answer:
NH4 CNO arrow CO(NH2)2 is the type of rearrangement reaction.

Question 15.
Give an example of double decomposition type of reaction.
Answer:
AgNO3 + NaCl = AgCl + NaNO3 is an example of double decomposition type of reaction.

Question 16.
State a limitation of a chemical equation.
Answer:
A reaction whether exothermic or endothermic is not known from a chemical equation.

Question 17.
Give an example of addition reaction.
Answer:
Carbon monoxide reacts with chlorine to form carbonyl chloride (COCl2) is an example of addition reaction.

Question 18.
Give an example of a substitution reaction.
Answer:
Chlorine reacts with methane (CH4) to form successively CH3 Cl, CH2 Cl2, CHCl3 and ultimately CCl4 in presence of sunlight is an example of a substitution reaction.

WBBSE Class 10 Physical Science Solutions Chapter 3 Chemical Calculations 2

Question 19.
Give an example of a thermal decomposition type of reaction.
Answer:
Calcium carbonate on heating decomposes into calcium oxide and carbon dioxide is an example of a thermal decomposition type of reaction.

Question 20.
Give an example of redox type of reaction.
Answer:
WBBSE Class 10 Physical Science Solutions Chapter 3 Chemical Calculations 3
is an example of redox type of reaction.

Question 21.
Balance the following reaction: Fe + H2 O(Fe3 O4 + H2.
Answer:
Balanced equation of Fe + H2 O arrow Fe3 O4 + H2 is
3 Fe + 4 H2 O = Fe3 O4 + 4 H2

Question 22.
2 SO2 + O2 ⇌ 2 SO3; what type of reaction is it ?
Answer:
2 SO2 + O2 ⇌ 2 SO3 is an example of reversible type of reaction.

WBBSE Class 10 Physical Science Solutions Chapter 3 Chemical Calculations

Question 23.
A + B2+ = A2+ + B; what type of reaction is it ?
Answer:
A + B2+ = A2+ + B is an example of oxidation-reduction type of reaction.

Question 24.
Balance the following reaction :
MnO2 + HCI → MnCI2 + CI2 + H2O
Answer:
Balanced equation of MnO2 + HCl → MnCl2 + Cl2 + H2O is
MnO2 + 4 HCl = MnCl2 + Cl2 + 2 H2O

Question 25.
Balance the following reaction :
Ag2 SO4 + BaCI2 → BaSO4 + AgCl.
Answer:
Balanced equation of Ag2 SO4 + BaCl2 → BaSO4 + AgCl is
Ag2 SO4 + BaCl2 = BaSO4 + 2 AgCl

Question 26.
Write the equation relating energy with mass.
Answer:
E = m c2

Question 27.
Write the relation between molecular mass and vapour density.
Answer:
Molecular mass = 2 × Vapour density.

WBBSE Class 10 Physical Science Solutions Chapter 3 Chemical Calculations

Question 28.
Calculate the vapour density of oxygen relative to hydrogen.
Answer:
Vapour density of oxygen
= \(\frac{Molecular mass of oxygen}{2} = [latex]\frac{32}{2}\) = 16

Question 29.
6 g carbon on complete combustion produce how much volume of CO2 at STP ?
Answer:
WBBSE Class 10 Physical Science Solutions Chapter 3 Chemical Calculations 4
12g carbon on complete combustion produce 22.4 L CO2 at STP
6g carbon on camplete combustion produce 11.2 L CO2 at STP

Question 30.
One mole of zinc reacts with excess H2 SO, to produce how many moles of hydrogen ?
Answer:
1 mole of hydrogen.

Short Answer type Questions :

Question 1.
What is empirical formula?
Answer:
Empirical formula : It is the formula of a compound which gives the simple whole number ratio of the atoms of various elements present in one molecule of compound.

Question 2.
What is molecular formula ?
Answer:
Molecular formula : It is the formula of a compound which gives the actual number of atoms of various elements present in one molecule of the compound.

Question 3.
What are the limitations of chemical formula ?
Answer:
Limitations of chemical formula :

  1. It fails to convey whether the elements in a molecule are present in the form of atoms or ions.
  2. It does not tell anything about the binding force that holds atoms in a molecule together.
  3. It does not tell us about the arrangement of various atoms with respect to one another within the molecule.

WBBSE Class 10 Physical Science Solutions Chapter 3 Chemical Calculations

Question 4.
What is structural formula ?
Answer:
Structural formula : A formula which gives the actual arrangement of the different atoms in the molecule or show how the different atoms in the molecule are linked together is called a structural or a graphic formula of the compound.

Question 5.
What is percentage of an element?
Answer:
Percentage of an element : Percentage of an element in a chemical compound is the number of parts by weight of it present in 100 parts by weight of the compound.

Question 6.
What are mass-mass relationship problems ?
Answer:
Mass-mass relation hip problems : In this type of problems, mass of one of the reactants/products is to be calculated if that of the other reactants/ products is to be calculated if that of the other reactants/ products are given.

Question 7.
What are mass-volume relationship problems ?
Answer:
Mass-volume relationship problems : In this type of problems, mass or volume of one of the reactants or products is calculated from the volume or mass of other substances.

Question 8.
What are volume-volume relationship problems ?
Answer:
Volume-volume relationship problems : In this type of problems, the volume of one of the reactants or products is to be calculated from the volume of some other reactant and product.

Question 9.
What is limiting reactant ?
Answer:
Limiting reactant : The reactant which is completely used and determines the amount of product formed is known as limiting reactant.

WBBSE Class 10 Physical Science Solutions Chapter 3 Chemical Calculations

Question 10.
What is theoretical yield ?
Answer:
Theoretical yield : The theoretical yield of a product is the maximum yield obtainable as calculated on the basis of the amount of limiting reactant used.

Question 11.
What is percent yield?
Answer:
Percent yield : Percent yield which is the ratio of the actual yield to the theoretical yield multiplied by 100 .

Question 12.
What do you mean by chemical reaction ?
Answer:
Chemical reaction: Any chemical change in matter which involves the transformation of matter into a new substance or new substances is termed as chemical reaction.

Question 13.
Define reactants in a chemical reaction.
Answer:
Reactants: The substances with which a chemical reaction is started are called the reactants.

Question 14.
Define products in a chemical reaction.
Answer:
Products: The substances formed as the result of the chemical reaction are called the products.

WBBSE Class 10 Physical Science Solutions Chapter 3 Chemical Calculations

Question 15.
Define direct combination reaction.
Answer:
Direct combination : The reaction in which two or more reactions combine directly forming molecules of new substance is called direct combination.
Example : Burning magnesium wire reacts directly with oxygen forming new molecules of magnesium oxide.
2 Mg + O2 = 2 MgO

Question 16.
What is meant by decomposition reaction?
Answer:
Decomposition : When molecule of a substance is broken down or decomposed in a chemical reaction to form two (or more) new substances in presence of heat or electricity. It is called decomposition.
Example : (a) Calcium carbonate on heating decomposes into calcium oxide and carbon dioxide.

WBBSE Class 10 Physical Science Solutions Chapter 3 Chemical Calculations 5

(b) Electricity is passed through acidulated water hydrogen and oxygen are produced due to decomposition of water.
2H2O = 2H2 + O2

Question 17.
Define double decomposition reaction.
Answer:
Double decomposition : The reaction in which the constituents of the molecules of reactants Change their position and forming molecules of new substances is called double decomposition.
Example : AgNO3 + NaCl = AgCl↓+ NaNO3

Question 18.
What is meant by acid-base reaction or neutralisation reaction ?
Answer:
Acid-base reaction or neutralisation reaction : In this reaction an acid reacts with a base forming salt and water. Equivalent amount of an acid neutralises an equivalent amount of a base. This is known as neutralisation reaction.

WBBSE Class 10 Physical Science Solutions Chapter 3 Chemical Calculations 6

Question 19.
Define oxidation-reduction type reaction.
Answer:
Oxidation-reduction reaction : The chemical reaction in which a chemical species loses electron(s) is an oxidation reaction and the reaction in which a chemical species gains electron(s) is called oxidation-reduction reaction.

WBBSE Class 10 Physical Science Solutions Chapter 3 Chemical Calculations 7

Question 20.
What is meant by addition reaction?
Answer:
Addition reaction: The reaction in which one reactant molecule directly combines with other molecule of reactant forming new molecules of product without leaving any part of molecules of reactants is known as addition reaction.

WBBSE Class 10 Physical Science Solutions Chapter 3 Chemical Calculations 8

Question 21.
Define substitution reaction.
Answer:
Substitution reaction: It is a chemical reaction in which one atom or molecule of a compound is replaced by another atom or molecule of another substance.
Example : Zn + CuSO4 = ZnSO4 + Cu

Question 22.
What is meant by rearrangement reaction?
Answer:
Rearrangement reaction: A reaction where a compound changes by internal arrangement of its atoms into another substance with different properties but having the same composition is known as rearrangement reaction.
Example : Ammonium cyanate changes to urea on heating.

WBBSE Class 10 Physical Science Solutions Chapter 3 Chemical Calculations 9

Question 23.
Define chemical equation.
Answer:
Chemical equation : A brief representation of a chemical reaction by using symbols of atoms of the elements and formulae of molecules of reactants and products maintaining the balance in between is known as chemical equation. It reveals both the qualitative ancL quantitative aspects of a chemical change.

Broad Answer Type Questions :

Question 1.
What steps should be followed to solve a problem based on chemical equation by weight-weight method ?
Answer:
Following steps are followed :

  1. First write down the complete balanced chemical equation.
  2. The molecular weight of the reactants and products are to be calculated from their respective formulae by adding atomic weights of the concerned elements.
  3. The unknown weight of the substance asked for in the question is to be calculated from the calculated weight in the equation.
  4. Same units are to be used for all quantities.

WBBSE Class 10 Physical Science Solutions Chapter 3 Chemical Calculations

Question 2.
What are the informations avallable from the formula CaCO3 ?
Answer:
The formula CaCO3 conveys the following informations :

  1. It stands for calcium carbonate.
  2. It tells that calcium carbonate is composed of calcium, carbon and oxygen elements.
  3. It represents that a molecule of calcium carbonate.
  4. It represents that a molecule of calcium carbonate is made up of One atom each of calcium and carbon and three atoms of oxygen.
  5. It stands for 100 parts by weight of calcium carbonate i.e. its molecular weight and signifies that the ratio of calcium, carbon and oxygen by weight in it is 40: 12: 48.

Question 3.
What is the method of writing chemical equation ?
Answer:
Method of writing chemical equation :
i. Initially the symbols of atoms of elements and formulae of molecules of reactants and products are written.

ii. Symbols and formulae of the reactants are written in the left side and these for products are in the right side. An arrow mark (→) is placed between reactants and products. For more than one reactants and products a plus sign (+) is to be given between the reactants and also between the products.

iii. Number of atoms of the reactants in the left hand side must be equal to the number of atoms of the products in the right hand side so as to maintain the law of conservation of mass. To equalise both sides, proper multiplication is required so that the number of atoms in both sides are same.

iv. Now the equation is expressed by replacing the arrow sign by a sign of equal ( = ).

WBBSE Class 10 Physical Science Solutions Chapter 3 Chemical Calculations

Question 4.
What informations are obtained from the chemical equations ?
Answer:
A chemical equation gives the following informations :
(a) Qualitative informations :

  • From chemical equation, the naming of elements and compounds taking part in the reaction and also the products are known.
  • From chemical equation, the symbols and the formulae of the reactants and products are also known.

(b) Quantitative informations :

  • Number of atoms or molecules of the reactants and products involved in the reaction are known.
  • How many parts of elements and compounds taking part in the reaction and producing are also known.
  • If the reactants and products are all gaseous then at the same temperature and pressure the ratio in volumes is known.

Question 5.
What informations are obtained from the equation :
C + O2 = CO2
Answer:
The following informations are obtained from the equation :
C + O2 = CO2

(a) Qualitative informations :
i. From the above equation it is known that carbon combines with oxygen producing carbon dioxide.
ii. Their respective symbols and formulae are also known.

(b) Quantitative informations :
i. One carbon atom combines with one oxygen molecule producing one carbon dioxide molecule.
Hence total number of atoms in the left hand side is (1 + 2) = 3 and the number of atoms in the right hand side is also 3.
ii. One gram-atom carbon reacts with one gram-molecule of oxygen giving rise one gram-molecule of carbon dioxide.
12 g carbon combines with 32 g oxygen yield 44 g carbon dioxide.
Hence, total mass in the left hand side (12 + 32) g = 44 g is equal to the total mass in the right hand side 44 g. It proves law of conservation of mass.
iii. Again 12 g carbon combines with 32 g oxygen producing 22.4 litre of carbon dioxide at STP.

WBBSE Class 10 Physical Science Solutions Chapter 3 Chemical Calculations

Question 6.
What informations are obtained from the equation :
2 H2 + O2 = 2 H2O
Answer:
The following informations are obtained from the equation :
2 H2O + O2 = 2 H2O

(a) Qualitative informations : The reactants are hydrogen and oxygen, water is the product.
(b) Quantitative informations :

  1. Two molecules of hydrogen combine with one molecule oxygen to produce two molecules water.
  2. Two volumes hydrogen and one volume oxygen combine to produce two volumes water at the same temperature and pressure and their ratio is 2 : 1 : 2
  3. Two moles of hydrogen and one mole oxygen combine chemically to produce two moles water.

Question 7.
What informations are obtained from the equation :
3 H2 + N2 = 2 NH3
Answer:
The following informations are obtained from the equation :
3 H2 + N2 = 2 N H3
(a) Qualitative informations : The reactants are hydrogen and nitrogen, ammonia is the product.
(b) Quantitative informations :

  1. Three molecules of hydrogen combine with one molecule nitrogen to produce two molecules ammonia.
  2. Three volumes hydrogen and one volume nitrogen combine to produce two volumes ammonia at the same temperature and pressure and their ratio is 3 : 1 : 2.
  3. Three moles of hydrogen and one mole nitrogen combine chemically to produce two moles ammonia.

Question 8.
What are the limitations of a chemical equation ?
Answer:
Limitations of a chemical equation :

  1. Under what conditions a chemical reaction is occurred i.e. pressure, temperature, catalyst are not known.
  2. A reaction whether exothermic or endothermic is not known from chemical equation.
  3. Chemical equation cannot provide information about completion of the reaction or attainment of equilibrium.
  4. A chemical equation does not give any idea about the reversibility of the reaction.
  5. From a chemical equation, the time required for the completion of the equation is not known.
  6. A reaction whether slow or fast does not know from the equation.
  7. Equation does not provide information regarding the nature of the reactants and products viz. solid, liquid or gas.

WBBSE Class 10 Physical Science Solutions Chapter 3 Chemical Calculations

Question 9.
What is meant by ‘balancing’ of a chemical equation ? Why it is necessary?
Answer:
Balancing of a chemical equation : It means making both the sides of the equation equal in respect of the kind and number of the atoms of the elements involved.
Balancing of a chemical equation is necessary : Due to the conservation of mass and indestructibility of matter, no atoms can be created or destroyed in a chemical reaction. Hence, the same kind of atoms in same numbers must be present in both the sides of the chemical equation.

Question 10.
Explain the method of balancing a chemical equation by trial and error method.
Answer:
Balancing of a chemical equation by trial and error method :
(i) In this method the proper numbers are putting before reactants and products so that the number of atoms in both sides are equal.
(ii) In doing so, we must cautious to see that all the reactants and products exist as molecules, and not as atoms, as free existence of atoms is not possible.

Question 11.
Balance the equation KClO3 ⟶ KCl + O2 by trial and error method.
Answer:
KClO3 ⟶ KCl + O2; it is seen that the number of atoms of oxyen in left hand side is 3 whereas in the right hand side it is 2. To equalize the number of atoms of oxygen in both sides it requires to multiply KClO3 by 2 and oxygen by 3.
2 KClO3 ⟶ KCl + 3 O2. So as to balance this equation if KCl is multiplied by 2 then the equation is properly balanced.
2 KClO3 = 2 KCl + 3 O2

WBBSE Class 10 Physical Science Solutions Chapter 3 Chemical Calculations

Question 12.
Write the following as balanced equations.
(i) Cu + HNO3 ⟶ Cu(NO3)2 + NO + H2 O
(ii) Pb(NO3)2 ⟶ PbO + NO2 + O2
(iii) Al + NaOH + H2 O ⟶ NaAlO2 + H2
(iv) CuO + NH3 ⟶ Cu + N2 + H2 O
Answer:
Balanced equations are :
(i) 3 Cu + 8 HNO3 = 3 Cu(NO3)2 + 2NO + 4H2 O
(ii) 2 Pb(NO3)2 = 2 PbO + 4 NO2 + O2
(iii) 2 Al + 2 NaOH + 2 H2 O = 2 NaAlO2 + 3H2
(iv) 2 NH3 + 3 CuO = 3 Cu + 2 N2 + 3H2O

Numerical problems :

Working formula :

(i) Molecular weight (M) = n × Emperical formula
(ii) Atomic weight of a gaseous element = \(\frac{\text { Actual yield }}{\text { Theoretical yield }} \times 100\)
Problems related to mass-mass relationship

Example 1:
How many grams of oxygen evolve when 122.5 g potassium chlorate is heated ? (Given, K = 39, Cl = 35.5, O = 16)
Answer:
The balanced equation is :
2 KClO = 2 KCl + 3 O2
2[39 + 35.5 + 3 × 16] g & 3[2 × 16] g
= 245 g = 96 g
By heating 245 g KClO3 96 g O2 is obtained
∴ By heating 122.5 g KClO3 \(\frac{96 \times 122.5}{245}\) = 48 g O2 is obtained

Example 2:
2.6 g zinc is treated with excess dil H2 SO4; how many gram oxygen combines with the evolved hydrogen?
Answer:
The balanced equation for production of hydrogen is :
Zn + H2 SO4 = ZnSO4 + H2
65g (2 × 1) g = 2g
So, 65 g zinc produces 2 g hydrogen
∴ 2-6 zinc produces \(\frac{2 \times 2.6}{65} \mathrm{~g}\) = 0.089 g hydrogen
Now, the reaction where hydrogen and oxygen combine is :
2 H2 + O2 = 2 H2 O
So, 4 g hydrogen combine with 32 g oxygen 0.08 g hydrogen combine with \(\frac{32 \times 0.08}{4} \mathrm{~g}\) = 0.64 g oxygen

WBBSE Class 10 Physical Science Solutions Chapter 3 Chemical Calculations

Example 3:
What is the observed loss in weight of 5 g calcium carbonate when it undergoes thermal decomposition?
Answer:
The balanced equation is :
CaCO3 = CaO + CO2
(40 + 12 + 16 × 3)g (12 + 16 × 2)g
= 100g = 44g
Loss in weight in the weight of CO2 that escapes
100 g CaCO3 produces 44g CO2
∴ 5 g CaCO3 produces \(\frac{44 \times 5}{1000}\) g = 2.2g CO2

Example 4:
What is the percentage of ammonia in that quantity of ammonium chloride that can produce 5g ammonia?
Answer:
The balanced equation of reaction is :
2 NH4 Cl + CaO = 2 NH3 + CaCl2 + H2 O
2[14 + 1 × 4 + 35.5] g 2[14 + l × 3] g
= 107 g = 34 g
So, 34 g ammonia is obtained form 107 g NH4 Cl

Example 5:
What weight of potassium chlorate of 96% purity will yield 4.8 g oxygen on complete thermal decomposition?
(Given, K = 39, Cl = 35.5, O = 16)
Answer:
The balanced equation is :
2 KClO3 = 2 KCl + 3 O2
245 g 96 g
∴ 96 g oxygen is obtained from 245 g KClO2 of 100% purity
∴ 4.8 g oxygen is obtained from \(\frac{245 \times 4 \cdot 8}{96}\) g
= 12.25 g KClO3 of 100% purity.
Let x gram of 96% purity contains 12.25 g KClO3 of 100% purity
∴ \(\frac{96}{100}\).x = 12.56
or, x = \(\frac{12 \cdot 25 \times 100}{96}\) = 12.76 (approx)
So, the required quantity of KClO3 = 12.76g

WBBSE Class 10 Physical Science Solutions Chapter 3 Chemical Calculations

Example 6:
On strong heating limestone decomposes into quicklime and carbon dioxide. How much quantity of limestone will produced on complete decomposition, 30 g of quicklime by the above reaction?
Answer:
The balanced equation is :
CaCO3 = CaO + CO2
[40 + 12 + 3 × 16] g [40 + 16] g
= 100 g = 56 g
So, 56 g CaO is obtained by the complete decomposition of 100 g CaCO3
∴ 30 g CaO is obtained by the complete decomposition of
\(\frac{100 \times 30}{56}\)g = 53.6 g CaCO3
Thus 53.6 g limestone will have to be decomposed.

Example 7:
How many grams of magnesium metal will give 1.2 g hydrogen on complete reaction with dilute H2 SO4 (Mg = 24, H = 1)
Answer:
The balanced equation is :
Mg + H2 SO4 = MgSO4 + H2
24 g (1 × 2) g = 2 g
So, 2 g H2 is obtained from 24 g Mg
∴ 1.2 g H2 is obtained from \(\frac{24 \times 1.2}{2}\)g Mg = 14.4 g Mg
So, 14.4 g Mg will be required.
Problems based on mass-volume relationship

WBBSE Class 10 Physical Science Solutions Chapter 3 Chemical Calculations

Example 8:
What volume of carbon dioxide measured at 300 K and 720 mm pressure be obtained on treatment of 1 g CaCO3 with dilute HCl ?
Answer:
The balanced equation is :
CaCO3 + 2 HCl = CaCl2 + H2O + CO2
1 mol(40 + 12 + 3 × 16) g        1 mol (22.4 lit at NTP)
= 100 g
So, 100g CaCO3 gives 22.4 lit CO2 at NTP
1g CaCO3 gives 0.224 lit CO2 at NTP

Reducing the volume to the given condition applying gas equation,

WBBSE Class 10 Physical Science Solutions Chapter 3 Chemical Calculations 10

P1 = 760 mm
V1 = 0.224 lit
T2 = 273 K
P2 = 720 mm
V1 = ?
T2 = 300 K

WBBSE Class 10 Physical Science Solutions Chapter 3 Chemical Calculations 11

= 0-2598 lit = 259-8 ml
Hence, 259.8 ml CO2 will be obtained at 300 K and 720 mm pressure on treatment of 1 g CaCO3 with HC 1.

Example 9:
How much potassium nitrate should be heated to get enough oxygen required to completely burn 56 lit of hydrogen at NTP ?
Answer:
The balanced equation is :

WBBSE Class 10 Physical Science Solutions Chapter 3 Chemical Calculations 12

= 2 × 22-4 lit
= 44.8 lit (at NTP)
So, 44.8 lit H2 requires 22.4 lit O2 at NTP
56 lit H, requires \(\frac{2 \cdot 24 \times 56}{44 \cdot 8}\) at NTP
= 28 lit O2 at NTP

The other balanced equation is :
WBBSE Class 10 Physical Science Solutions Chapter 3 Chemical Calculations 13
= 202 g
Now, 22.4 lit O2 at NTP are obtained by heating 202 g KNO3
∴ 28 lit O2 at NTP are obtained by heating \(\frac{202}{22.4}\) × 28 g KNO3
= 252.5 g KNO3

WBBSE Class 10 Physical Science Solutions Chapter 3 Chemical Calculations

Problems based on volume-volume relationship.

Example 10:
In the Ostwald process for the manufacture of nitric acid, ammonia gas is burnt in oxygen in the presence of a pt-catalyst. What volume of O2 is required and what volume of NO is formed in the combustion of 500 lit of NH3.
Answer:
The balanced equation is :
WBBSE Class 10 Physical Science Solutions Chapter 3 Chemical Calculations 14

(i) 4 lit of NH3 requires 5 lit of O2 for combustion
500 lit of NH3 requires \(\frac{5 \times 500}{4}\) lit = 625 lit O2
(ii) 4 lit of NH3 produces 4 lit NO
∴ 500 lit of NH3 produces \(\frac{4}{4}\) × 500 lit NO = 500 lit NO

Example 11:
Calculate the volume of oxygen necessary to burn completely 5 lit butane gas. What is the volume of carbon dioxide formed?
Answer:
The balanced equation is :
WBBSE Class 10 Physical Science Solutions Chapter 3 Chemical Calculations 15

So, 2 lit of butane at NTP requires 13 lit O2 at NTP
∴ 5 lit of butane at NTP requires \(\frac{13 \times 5}{2}\) lit = 32.5 lit of O2 at NTP.
Also 2 lit of butane at NTP produce 8 lit CO2 at NTP
∴ 5 lit of butane of NTP produce
= \(\frac{8}{2}\) × 5 lit = 20 lit CO2 at NTP

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