9th Class Chemistry · Punjab Board · Chapter 4

Chapter 4: StoichiometryNumericals Solved Step by Step

All 14 numericals of 9th Class Chemistry Chapter 4, Stoichiometry, solved step by step: Given Data, To Find, Formula, Calculation and Result, the way you write them in the exam. Punjab Board new syllabus.

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Chapter 4· Stoichiometry

Numericals Solved Step by Step

Each numerical in the order you write it in the exam: Given Data, To Find, Formula, Calculation, Result.

Example 4.1 (book)

25 g of limestone (CaCO₃) reacts with an excess of hydrochloric acid: CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂. How much calcium chloride (CaCl₂) will be produced?

Given Data

Mass of CaCO₃ = 25 g
Molar mass of CaCO₃ = 100 g mol⁻¹
Molar mass of CaCl₂ = 111 g mol⁻¹

To Find

Mass of CaCl₂

Formula

1 mol CaCO₃ (100 g) → 1 mol CaCl₂ (111 g)

Calculation

100 g of limestone produces calcium chloride = 111 g.
1 g of limestone produces 111 ÷ 100 g.
25 g of limestone produces 111 ÷ 100 × 25 = 27.75 g.

Result

27.75 g of CaCl₂

Example 4.2 (book)

1.80 moles of ethyl alcohol, when burnt in air completely, will utilize how many moles of oxygen gas? Also calculate the number of moles of CO₂ produced.

Given Data

Moles of ethyl alcohol = 1.80 mol
C₂H₅OH(l) + 3O₂(g) → 2CO₂(g) + 3H₂O(g)

To Find

Moles of O₂ used
Moles of CO₂ produced

Formula

mole ratio C₂H₅OH : O₂ : CO₂ = 1 : 3 : 2

Calculation

One mole of ethyl alcohol needs 3 moles of O₂.
1.8 moles need 3 ÷ 1 × 1.8 = 5.4 moles O₂.
One mole of ethyl alcohol produces 2 moles of CO₂.
1.8 moles produce 2 ÷ 1 × 1.8 = 3.6 moles CO₂.

Result

5.4 mol O₂ used; 3.6 mol CO₂ produced

Example 4.3 (book)

Aluminium metal reacts with oxygen to produce aluminium oxide. How many grams of oxygen will be required to react completely with 0.3 moles of aluminium?

Given Data

Moles of Al = 0.3 mol
4Al(s) + 3O₂(g) → 2Al₂O₃(s)
Molar mass of O₂ = 32 g mol⁻¹

To Find

Mass of O₂

Formulas

mole ratio Al : O₂ = 4 : 3

m = n × M

Calculation

4 moles of Al need 3.0 moles of O₂.
1.0 mole of Al needs ¾ mole.
0.3 mole of Al needs ¾ × 0.3 = 0.225 mole O₂.
Mass = 32 × 0.225 = 7.2 g.

Result

7.2 g of O₂

Note: The book's last line says "will have molar mass"; it means mass. The molar mass of O₂ is always 32 g mol⁻¹.

Example 4.4 (book)

How many molecules of water will be produced if we react 5 g of hydrogen gas with excess of oxygen gas?

Given Data

Mass of H₂ = 5 g
2H₂(g) + O₂(g) → 2H₂O(l): 4 g, 32 g, 36 g

To Find

Number of water molecules

Formulas

4 g H₂ → 36 g H₂O

N = (m ÷ M) × Nᴀ

Calculation

4 g of hydrogen produce 36 g of H₂O.
5 g produce 36 ÷ 4 × 5 = 45 g of H₂O.
18 g (1 mole) of H₂O contains 6.022 × 10²³ molecules, so 36 g contains 12.04 × 10²³.
45 g contains 45 ÷ 36 × 12.04 × 10²³ = 1.505 × 10²⁴ molecules.

Result

1.505 × 10²⁴ molecules of water

Sample problem: molecular formula

The empirical formula of a compound is CH. Its molecular mass is 78 g mol⁻¹. Find its molecular formula.

Given Data

Empirical formula = CH
Molecular mass = 78 g mol⁻¹

To Find

Molecular formula

Formulas

n = molar mass ÷ empirical formula mass

Molecular formula = n (empirical formula)

Calculation

Empirical formula mass of CH = 12 + 1 = 13.
n = 78 ÷ 13 = 6.
Molecular formula = (CH)₆ = C₆H₆.

Result

C₆H₆ (benzene)

In-text exercise 1

The empirical formula of a compound is CH₂O. Its molar mass is 180 g mol⁻¹. Determine its molecular formula.

Given Data

Empirical formula = CH₂O
Molar mass = 180 g mol⁻¹

To Find

Molecular formula

Formula

n = molar mass ÷ empirical formula mass

Calculation

Empirical formula mass = 12 + 2(1) + 16 = 30.
n = 180 ÷ 30 = 6.
Molecular formula = (CH₂O)₆ = C₆H₁₂O₆.

Result

C₆H₁₂O₆ (glucose)

In-text exercise 2

The empirical formula of a compound is CH₂O. Its molar mass is 60 g mol⁻¹. Determine its molecular formula.

Given Data

Empirical formula = CH₂O
Molar mass = 60 g mol⁻¹

To Find

Molecular formula

Formula

n = molar mass ÷ empirical formula mass

Calculation

Empirical formula mass = 30.
n = 60 ÷ 30 = 2.
Molecular formula = (CH₂O)₂ = C₂H₄O₂.

Result

C₂H₄O₂ (acetic acid)

Exercise 4(iii)

How many grams of CO₂ will be produced when we react 10 g of CH₄ with excess of O₂ according to the equation CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g)?

Given Data

Mass of CH₄ = 10 g
Molar mass CH₄ = 16 g mol⁻¹
Molar mass CO₂ = 44 g mol⁻¹

To Find

Mass of CO₂

Formula

1 mol CH₄ (16 g) → 1 mol CO₂ (44 g)

Calculation

16 g of CH₄ produce 44 g of CO₂.
1 g produces 44 ÷ 16 g.
10 g produce 44 ÷ 16 × 10 = 27.5 g.

Result

27.5 g of CO₂

Exercise 4(iv)

How many moles of coal are needed to produce 10 moles of CO according to the equation 3C(s) + O₂(g) + H₂O(l) → H₂(g) + 3CO(g)?

Given Data

Moles of CO = 10 mol
Mole ratio C : CO = 3 : 3

To Find

Moles of coal (C)

Formula

3 mol C → 3 mol CO

Calculation

3 moles of C produce 3 moles of CO, i.e. 1 : 1.
10 moles of CO need 3 ÷ 3 × 10 = 10 moles of C.

Result

10 mol of coal (120 g of carbon)

Exercise 4(v)

How much SO₂ is needed in grams to produce 10 moles of sulphur? 2H₂S(g) + SO₂(g) → 2H₂O(l) + 3S(s)

Given Data

Moles of S = 10 mol
Molar mass SO₂ = 32 + 32 = 64 g mol⁻¹

To Find

Mass of SO₂

Formulas

1 mol SO₂ → 3 mol S

m = n × M

Calculation

3 moles of S need 1 mole of SO₂.
10 moles of S need 1 ÷ 3 × 10 = 3.33 moles of SO₂.
Mass = 3.33 × 64 = 213.3 g.

Result

213.3 g of SO₂

Exercise 4(vi)

How much ammonia is needed in grams to produce 1 kg of urea fertilizer? 2NH₃ + CO₂ → CO(NH₂)₂ + H₂O

Given Data

Mass of urea = 1 kg = 1000 g
Molar mass urea CO(NH₂)₂ = 12 + 16 + 2(14 + 2) = 60 g mol⁻¹
Molar mass NH₃ = 17 g mol⁻¹

To Find

Mass of NH₃

Formula

2 mol NH₃ (34 g) → 1 mol urea (60 g)

Calculation

Moles of urea = 1000 ÷ 60 = 16.67 mol.
Moles of NH₃ = 2 × 16.67 = 33.33 mol.
Mass of NH₃ = 33.33 × 17 = 566.7 g.

Result

566.7 g of NH₃ (about 0.567 kg)

Exercise 4(vii)

Calculate the number of atoms in (a) 3 g of H₂ (b) 3.4 moles of N₂ (c) 10 g of C₆H₁₂O₆.

Given Data

M(H₂) = 2 g mol⁻¹; M(C₆H₁₂O₆) = 180 g mol⁻¹
Nᴀ = 6.022 × 10²³

To Find

Number of atoms in each

Formulas

n = m ÷ M

atoms = n × Nᴀ × atoms per molecule

Calculation

(a) n = 3 ÷ 2 = 1.5 mol H₂; atoms = 1.5 × 6.022 × 10²³ × 2 = 1.807 × 10²⁴.
(b) atoms = 3.4 × 6.022 × 10²³ × 2 = 4.095 × 10²⁴.
(c) n = 10 ÷ 180 = 0.0556 mol; molecules = 3.346 × 10²²; each has 6 + 12 + 6 = 24 atoms; atoms = 24 × 3.346 × 10²² = 8.03 × 10²³.

Result

(a) 1.807 × 10²⁴ (b) 4.095 × 10²⁴ (c) 8.03 × 10²³ atoms

Exercise 3(iv) (CRQ)

When 8.657 g of a compound were converted into elements, it gave 5.217 g of carbon, 0.962 g of hydrogen and 2.478 g of oxygen. Calculate the percentage of each element.

Given Data

Mass of compound = 8.657 g
C = 5.217 g, H = 0.962 g, O = 2.478 g

To Find

% C, % H, % O

Formula

% = mass of element ÷ mass of compound × 100

Calculation

% C = 5.217 ÷ 8.657 × 100 = 60.26 %.
% H = 0.962 ÷ 8.657 × 100 = 11.11 %.
% O = 2.478 ÷ 8.657 × 100 = 28.62 %.

Result

C 60.26 %, H 11.11 %, O 28.62 %

Investigative 5(i)

Drinking eight glasses of water a day is believed to keep one hydrated. If a glass holds 400 cm³ of water on average, how many moles of water are needed for a single adult?