Stoichiometry Chemistry Class 11 Notes Chapter 1 for kpk

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

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.

---

---

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 (0^oC 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,

---