Stoichiometry Chemistry Class 11 Notes Chapter 1 for kpk textbook boards Peshawar, Adam jee practical center notes, cbse, ncert, and federal board.
Fsc Part 1 notes for Chemistry Chapter 1 adamjee
Table of Contents
What is a gram atom? Why is the concept of gram atoms useful in chemistry?
Gram atom:
The relative atomic mass of the element described in grams is called 1 gram atom. For example, the relative atomic scale of chlorine (CL) is 35.5 amu and the ratio of sulfur (s) is 32.1. Therefore, one gram atom of chlorine will be 35.5 g and one gram atom of sulfur will be 32.1 g.
The importance of imagination
The idea of a gram atom is practical because it is unbelievable to look at an individual atom or weigh it or make a large number of atoms precise. While the gram atom (sesame) contains a large number of atoms, which we can easily see and weigh. Therefore, considering the atomic point of view of a gram, stoichiometric calculations are easily made.
Stoichiometry Chemistry Class 11 Notes
The mass of 5 moles of an element X is 60 g. Calculate the molar mass of this element. Name the element?
As Carbon has molar mass 12 g/mol, so the unknown element X is carbon.
Explain why balanced chemical equations are used in stoichiometric problems?
According to the law of protection of mass, the matter (mass) remains conserved. Therefore the mass of creation is always equal to the mass of reactants. A level chemical equation is used to account for this saving of mass in stoichiometric difficulties. For example,
How will you identify the limiting reagent in a reaction?
Answer:
Identification of Limiting Reagent:
“Limiting reagent in a chemical reaction is the one which is consumed earlier in the reaction and hence controls the amount of product formed”.
Following are the steps that help in identifying the limiting reagent.
- The amounts of reactants (if given in mass units) are first converted into moles.
- Using balanced chemical equations, the moles of the desired product is calculated from the available moles of each reactant.
- The reactant which gives the least number of moles of the required product will be the limiting reagent.
- Alternatively comparison of the moles of reactants, with the help of balanced chemical equation, also helps in identifying the limiting reagent.
Define the molar volume of a gas. What will be the volume of 2.5 moles of H2 gas and 60g of NH3 at STP?
Answer:
Molar Volume:
One mole of any gas at STP (0oC and 1 atm) occupies a volume of 22.4 dm3. This is called the molar volume of a gas.
For hydrogen,
1 mol of H2 gas = 22.4 dm3
2.5 moles of H2 gas = 2.5 x 22.4 = 56 dm3
For NH3,
1 mole (17 g) of NH3 = 22.414 dm3
1 g of NH3 = 22.4/17 dm3
60 g of NH3 = (22.4/17) x 60 = 79.05 dm3
So, the volume of 2.5 moles of H2 gas will be 56 dm3 and the volume of 60 g of NH3 will be 79.05 dm3 at STP.
Why the actual yield is usually less than the theoretical yield of a reaction?
Answer:
The actual yield is always less than the theoretical yield due to the following reasons.
- The reaction may have not gone to completion, i.e. all the amounts of the reactants may not have converted into products, due to the reversibility of the reaction.
- Some of the amounts of the reactants may have converted into some other product (not known) by a side reaction, parallel reaction or chain reaction. All the product may not be retrieved in the process due to partial loss during the experiment.
- Some material may be lost in carrying out the reaction or in transferring and separating the product. This is called mechanical loss of the product (Human error).
- Reaction conditions (like temperature, pressure, concentration etc.) might have been disturbed.
Long Questions
Q.3 a) What is formula mass of a compound? What are the steps involved in calculating the formula mass of a compound. Explain with an example.
Answer:
Formula Mass:
The formula mass of the compound is the sum of the atomic weights of individual atoms present in the molecule.
Steps to Calculate Formula Mass
Formula mass of the compound can be determined by the following steps.
- First of all, the formula of the substance is written. For example, formula of sulphuric acid is H2SO4.
- Relative formula masses of elements in substance are worked out. Sulphuric acid contains two hydrogen atoms (Atomic mass 1 amu), one Sulphur (atomic mass 32 amu) and four oxygen atoms (atomic mass 16 amu). So relative formula mass for sulphuric acid is;
M = (2×1) + (1×32) + (4×16) = 98 amu
Q.3 b) Calculate formula masses of the following compounds:
i) HNO3
ii) C6H12O6
iii) C3H8
iv) C2H5OH
v) Al2O3
vi) K2Cr2O7
Answer:
i) HNO3
As we know that,
H = 1 amu, N = 14 amu, O = 16 amu
Formula mass of HNO3 = (1 x 1) + (1 x 14) + (3 x 16) = 63 amu
ii) C6H12O6
As we know that,
C=12 amu, H=1 amu, O=16 amu
Formula mass of C6H12O6 = (6 x 12) + (12 x 1) + (6 x 16) = 180 amu
iii) C3H8
As we know that,
C=12 amu, H=1 amu
Formula mass of C3H8 = (3 x 12) + (8 x 1) = 44 amu
iv) C2H5OH
As we know that,
C=12 amu, H=1 amu, O=16 amu
Formual mass of C2H5OH = (2 x 12) + (5 x 1) + (1 x 16) + (1 x 1) = 56 amu
v) Al2O3
As we know that,
Al = 27 amu, O=16 amu
Formula mass of Al2O3 = (2 x 27) + (3 x 16) = 102 amu
vi) K2Cr2O7
As we know that,
K=39 amu, Cr=52 amu, O=16 amu
Formula mass of K2Cr2O7 = (2 x 39) + (2 x 52) + (7 x 16) = 294 amu
Q.4 a) Define and explain mole and Avogadro’s number with examples.
Answer:
Mole:
The amount of a chemical substance which contains 6.023 x 1023 particles (atoms, molecules or formula units) in it is called mole.
Mole is a counting unit just like a dozen (12 similar things) or a gross (144 similar things). It is difficult to deal with individual atoms or molecules because they have very small sizes. That is why a very large counting unit (Avogadro’s number 6.023 x 1023) is suitable to use.
Avogadro ’s number:
One mole of any substance contains 6.023 x 1023 particles (atoms, molecules or ions) in it. This constant number is called Avogadro’s number.
Examples
1 mole of sodium (Na) = 23 g of Na = 6.023 x 1023 atoms of Na
1 mole of water (H2O) = 18 g of H2O = 6.023 x 1023 molecules of H2O
Q.4 b) Given the equation.
CH4(g) + 2O2(g) → CO2(g) + 2H2O(g) + Heat.
How can this equation be read in terms of particles, moles and masses?
Answer:
In terms of particles, 6.023 x 1023 molecules (particles) of CH4 react with 2 x 6.023 x 1023 molecules of O2 and give 6.023 x 1023 molecules of CO2 and 2 x 6.023×1023 molecules of H2O.
In terms of moles, one mole of CH4 reacts with two moles of O2 and gives one mole of CO2 and two moles of H2O.
In terms of masses, 16 grams of CH4 reacts with 64 grams of O2 and give 44 grams of CO2 and 36 grams of H2O. This is in accordance with the law of conservation of mass.
Q.5 a) What do you mean by percentage composition of a compound? How the percentage of an element is calculated in a compound.
Answer:
Percentage Composition:
The percentage by mass of elements present in a compound is called percentage composition. First, the elements present in a compound are identified, and the molecular mass or formula mass of the compound is determined, then the following formula is used to find the percentage compositions of elements.
or,
Q.5 b) Calculate the percentage composition of each of the following compounds, (given atomic weights of the elements).
i) MgSO4
ii) C3H6O
iii) KMnO4
iv) C6H6
v) NaAl(SO4)2
vi) CaCO3
vii) CH4
(Mg=24, S=32, O=16, C=12, K=39, Na=23,Mn=55, Ca=40, Al=27, H=1)
Answer:
i) MgSO4
From the formula, we get:
Molar mass of MgSO4 = (1×24) + (1×32) + (4×16) = 120 g/mol
Mass contributed by Mg = 24 g
Mass contributed by S = 32 g
Mass contributed by O = 64 g
Percentage of each element in MgSO4 is as follows,
ii) C3H6O
From the formula, we get:
Molar mass of C3H6O = (3×12) + (6×1) + (1×16) = 58 g/mol
Mass contributed by C = 36 g
Mass contributed by H = 6 g
Mass contributed by O = 16 g
Percentage of each element in C3H6O is as follows,
(iii) KMnO4
From the formula, we get:
Molar mass of KMnO4 = (1×39) + (1×55) + (4×16) = 158 g/mol
Mass contributed by K = 39 g
Mass contributed by Mn = 55 g
Mass contributed by O = 64 g
Percentage of each element in KMnO4 is as follows,
iv) C6H6
From the formula, we get:
Molar mass of C6H6 = (6×12) + (6×1) = 78 g/mol
Mass contributed by C = 72 g
Mass contributed by H = 6 g
Percentage of each element in C6H6 is as follows,
v) NaAl(SO4)2
From the formula, we get:
Molar mass of NaAl(SO4)2 = (1×23) + (1×27) + 2[(1×32) + (4×16)] = 242 g/mol
Mass contributed by Na = 23 g
Mass contributed by Al = 27 g
Mass contributed by S = 64 g
Mass contributed by O = 128 g
Percentage of each element in NaAl(SO4)2 is as follows,
vi) CaCO3
From the formula, we get:
Molar mass of CaCO3 = (1×40) + (1×12) + (3×16) = 58 g/mol
Mass contributed by C = 40 g
Mass contributed by H = 12 g
Mass contributed by O = 48 g
Percentage of each element in CaCO3 is as follows,
vii) CH4
From the formula, we get:
Molar mass of CH4 = (1×12) + (4×1) = 16 g/mol
Mass contributed by C = 12 g
Mass contributed by H = 4 g
Percentage of each element in CH4 is as follows,
Read more: Physics Class 11 Notes Thermodynamics Chapter 10 for kpk
Q.6 a) Differentiate between a “limiting reagent” and “a reagent in excess”. How will you identify the limiting reagent in a chemical reaction?
Answer:
Difference between Limiting reagent and Excess reagent:
| Limiting reagent | Excess reagent |
| The reagent which is consumed earlier in a chemical reaction and thus controls the amount of product is known a limiting reagent. | The reagent which is present in excess amount and some of its amount is left un-reacted at the end of the reaction is called excess reagent. |
| For example, when coal is burnt in air, combustion stops when coal (carbon) is consumed, thus coal is a limiting reactant while oxygen (air) is present in excess amount in this reaction |
Identification of Limiting Reagent:
Following are the steps that are used for the identification of the limiting reagent.
- The amounts of reactants (if given in mass units) are first converted into moles.
- Using balanced chemical equations, the moles of the desired product are calculated from the available moles of each reactant.
- The reactant which gives the least number of moles of the required product will be the limiting reagent.
- Alternatively comparison of the moles of reactants, with the help of a balanced chemical equation, also helps in identifying the limiting reagent.
Q.6 b) How does a limiting reagent control the amount of the product formed? Give an example.
Answer:
Limiting reagent is the reagent which is present in small amount stoichiometrically. Due to small amount, the limiting reagent is consumed earlier in a reaction and when it is consumed completely the reaction stops (although other reagents are present in the reaction). Hence limiting reagent controls the amount of product formed in a chemical reaction.
Example:
When coal is burnt in air, combustion stops when coal (carbon) is consumed completely although oxygen is still present. Thus coal (carbon) is limiting reagent in this reaction because it is controlling the amount of product formed in this combustion reaction.
Q.7) A technician weighs 40 g of sodium chloride. How many moles of formula units are in the sample?
Answer:
Mass of NaCl = 40 g
Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol
Moles of formula units of NaCl present in the sample are calculated as,
Hence, 0.683 moles are present in 40 g sample of sodium chloride.
Q.8) Calculate the mass in grams of
a) 7.75 moles of Al2O3
b) 15 moles of H2SO4
c) 1.0 x 1025 molecules of H2O.
Answer:
a) 7.75 moles of Al2O3
Moles of Al2O3 = 7.75 moles
Molar mass of Al2O3 = (2×27) + (3×16) = 102 g/mol
Using the formula,
b) 15 moles of H2SO4
Moles of H2SO4 = 15 moles
Molar mass of H2SO4 = (2×1) + (1×32) + (4×16) = 98 g/mol
Using the formula,
c) 1.0 x 1025 molecules of H2O
No. of molecules of H2O = 1.0 x 1025 molecules
Molar mass of H2O = (2×1) + (1×16) = 18 g/mol
Using the formula,
Q.10) Calculate the mass percentage of a metal in a compound that is formed by 0.233 g of metal combining with 0.354 g of oxygen.
Answer:
Mass of metal = 0.233 g
Mass of oxygen = 0.354 g
As, the metal reacts with oxygen completely and there is no reagent in excess, so the mass of compound is given as,
Mass of compound = 0.233 + 0.354 = 0.587 g
Using formula for percentage composition,
Q.11) Given the equation: 2H2(g) + O2(g) → 2H2O(g)
a) How many moles of water will be obtained by burning 5.6 moles of O2 in an excess of H2?
b) How many moles of O2 would be needed to react 58.5 g of H2 to form water?
c) How many grams of H2 would be needed to form 120 g of H2O?
Answer:
a)
Moles of O2 = 5.6 moles
As, H2 is present in excess, so the limiting reagent O2 will control the amount of product (water).
According to balanced chemical equation,
2 moles of O2 ≅ 2 moles of H2O
1 mole of O2 ≅ 1 mole of H2O
5.6 moles of O2 ≅ 5.6 moles of H2O
So, 5.6 moles of water will be obtained by burning 5.6 moles of O2 in excess H2.
b)
Mass of H2 = 58.5 g
Molar mass of H2 = 2.016 g/mol
According to balanced chemical equation,
2 moles of H2 ≅ 1 mol of O2
29.02 moles of H2 ≅ (1/2) x 29.02 = 14.5 moles
Hence, 14.5 moles of O2 will be needed to react completely with 58.5 g of H2 to form water.
c)
Mass of water = 120 g
Molar mass of water (H2O) = (2×1) + 16 = 18 g/mol
From balanced chemical equation,
2 moles of water ≅ 2 moles of H2
1 moles of water ≅ 1 moles of H2
6.67 moles of water ≅ 6.67 moles of H2
Mass of H2 = Moles of H2 x Molar mass of H2 = 6.67 x 2.016 = 13.45 g
Hence, 13.45 grams of H2 would be needed to form 120 g (6.67 moles) of water.
Q.12 Given the equation N2(g) + 3H2(g) → 2NH3(g) at STP. How many moles of NH3 would be formed if 6.3 dm3 of N2 gas react with an excess of H2?
Answer:
From the concept of molar volume, the given volume of N2 is converted into moles.
Volume of N2 gas at STP = 6.3 dm3
Molar volume = 22.4 dm3
From balanced chemical equation,
1 mole of N2 ≅ 2 moles of NH3
0.28 moles of N2 ≅ 2 x 0.28 = 0.56 moles of NH3
Hence, 0.56 moles of NH3 would be formed if 6.3 dm3 of N2 reacts with an excess of H2.
Q.13) Calculate the mass of Mg metal required to consume 2560g of CO2 in the reaction. 2Mg(s) +CO2(s) →2MgO(s) +C(s)
Answer:
Mass of CO2 = 2560 g
Molar mass of CO2 = 12 + 2(16) = 44 g/mol
From balanced chemical equation,
1 mol CO2 ≅ 2 moles Mg
58.18 moles CO2 ≅ 2 x 58.18 = 116.36 moles of Mg
Mass of Mg required = Moles of Mg x Molar mass of Mg
= 116.36 x 24 = 2793 g
Hence, 2793 grams of Mg metal is required to consume 2560 grams CO2 in the given reaction
Q.14) When steam is passed through red hot carbon, a mixture of H2 and CO gas, called water gas, is formed.
H2O(g) + C(s) → CO(g) + H2(g)
a) Which is the limiting reagent if 24.5 g of carbon is mixed with 1.89 moles of water vapours?
b) Calculate the amount (in grams) of the excess reagent left unreacted.
Answer:
a)
Mass of carbon = 24.5 g
Molar mass of carbon = 12 g/mol
From balanced chemical equation,
1 mole of C ≅ 1 mole of H2
2.04 moles C ≅ 2.04 moles H2
Hence, 24.5 g (2.04 moles) of carbon can produce 2.04 moles of H2 (product).
From balanced chemical equation,
1 mole H2O ≅ 1 mole H2
1.89 moles H2O ≅ 1.89 moles H2
Hence, 1.89 moles of water vapours (H2O) can produce 11.89 moles of H2 (product).
As, H2O (water vapours) produces less amount of product, so water is limiting reagent in this reaction and reaction will stop when water is consumed.
b)
1 mole H2O ≅ 1 mole C
1.89 moles H2O ≅ 1.89 moles C
Amount of C left unreacted = Initial moles of C – Moles of C reacted
= 2.04 – 1.89 = 0.15 moles of C
Mass of C = Moles of C x Molar mass of C = 0.15 x 12 =1.8 g
Hence, 1.8 grams of carbon (excess reagent) are left unreacted at the end of reaction.
Q.15) Calculate the percentage yield if 6.53 g of hydrogen gas is produced when 5 moles of zinc is consumed in the reaction:
Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g)
Answer:
Moles of Zn consumed = 5 moles
From the balanced chemical equation, theoretical yield (amount of H2) is calculated as;
1 mole Zn ≅ 1 mole H2
5 moles Zn ≅ 5 moles H2
Mass of Zn produced = Moles of H2 x Molar mass of H2
= 5 x 2.016 = 10.08 g
The amount of H2 (product) calculated above from the stoichiometric equation is the theoretical yield. So,
Theoretical yield = 10.08 g
Actual yield = 6.53 g
Q.16) The percentage yield of the following reaction is 85 %.
2 Al(s) + 3 CI2(g) → 2 AICI3(s)
How many grams of AICI3 will be actually obtained from 100 g of aluminium metal.
Answer:
In this problem, we have to calculate the actual yield of AlCl3 which can be manipulated from the given formula.
Theoretical yield is calculated from given stoichiometric equation.
Mass of Al metal = 100 g
Molar mass of Al = 27 g/mol
Molar mass of AlCl3 = 27 + 3(35.5) = 133.5 g/mol
From balanced chemical equation,
2 moles of Al ≅ 2 moles of AlCl3
3.70 moles Al ≅ 3.70 moles AlCl3
Mass of AlCl3 formed = Moles of AlCl3 x Molar mass of AlCl3
= 3.70 x 133.5 = 493.95 g
So,
Theoretical yield=493.5g
Using formula,
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