Chemical Equations

Chemical Equations

1. Reactants → Products

• Reactants are the starting substances in a chemical reaction.
• Products are the new substances formed after the reaction.

A chemical equation is written as:$$\textbf{Reactants} \rightarrow \textbf{Products}$$Reactants→Products

The arrow (→) shows the direction of the reaction.

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2. Important Historical Contributions

• 1774 – Joseph Priestley discovered Oxygen.
• 1789 – Antoine Lavoisier proposed the Law of Conservation of Mass.

These discoveries helped scientists understand how chemical reactions work.

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3. Balanced Chemical Equation

A balanced equation has:

👉 The same number of atoms of each element on both sides of the equation.

This follows the Law of Conservation of Mass, which states:

Matter is neither created nor destroyed during a chemical reaction.

So,$$\textbf{Mass of Reactants = Mass of Products}$$Mass of Reactants = Mass of Products

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4. Example

$$O_2 + 2H_2 \rightarrow 2H_2O$$O2​+2H2​→2H2​O

This equation shows:

• Oxygen reacts with Hydrogen
• Water is formed
• The equation is balanced because the number of Hydrogen and Oxygen atoms is equal on both sides.

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5. Coefficients

• Coefficients are the numbers written in front of chemical formulas.
• They are used to balance equations.
• Example: In 2H₂O, the number 2 is the coefficient.

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Quick Revision Points

✔ Reactants → Products
✔ Balanced equation = equal atoms on both sides
✔ Law of Conservation of Mass
✔ Coefficients help in balancing

Chemical Equations (3–4 Marks Answer)

A chemical equation is the symbolic representation of a chemical reaction using chemical formulas.

In a chemical equation:

• Reactants are the starting substances.
• Products are the substances formed.
• The arrow (→) shows the direction of the reaction.

Example:$$O_2 + 2H_2 \rightarrow 2H_2O$$O2​+2H2​→2H2​O

A balanced chemical equation has the same number of atoms of each element on both sides.

This follows the Law of Conservation of Mass (1789 – Lavoisier), which states that matter is neither created nor destroyed in a chemical reaction.

Coefficients are the numbers placed before chemical formulas to balance the equation.

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Mass

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🔢 Numerical Practice Questions | Download PDF

Q1. Balance the equation:

$$H_2 + O_2 \rightarrow H_2O$$H2​+O2​→H2​O

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Q2. Balance the equation:

$$Fe + O_2 \rightarrow Fe_2O_3$$Fe+O2​→Fe2​O3​

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Q3. Balance the equation:

$$Al + O_2 \rightarrow Al_2O_3$$Al+O2​→Al2​O3​

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Q4. If 4 g of Hydrogen reacts completely with Oxygen, how many grams of Oxygen are required?

(Given reaction: $2H_2 + O_2 \rightarrow 2H_2O$2H2​+O2​→2H2​O)
(Atomic mass: H = 1, O = 16)

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Q5. Calculate the mass of water formed when 32 g of Oxygen reacts completely with Hydrogen.

Reaction: $2H_2 + O_2 \rightarrow 2H_2O$2H2​+O2​→2H2​O

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Q6. In the reaction

$$C + O_2 \rightarrow CO_2$$C+O2​→CO2​

How many grams of CO₂ are formed when 12 g of Carbon burns completely?

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Q7. 56 g of Iron reacts with Oxygen to form Iron(III) oxide. Calculate the mass of Fe₂O₃ formed.

(Balanced reaction required first.)

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Q8. In the reaction

$$2Mg + O_2 \rightarrow 2MgO$$2Mg+O2​→2MgO

How many grams of MgO are formed from 24 g of Magnesium?
(Atomic mass: Mg = 24, O = 16)

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Q9. If 10 g of Calcium reacts with water:

$$Ca + 2H_2O \rightarrow Ca(OH)_2 + H_2$$Ca+2H2​O→Ca(OH)2​+H2​

Calculate the mass of Hydrogen gas produced.
(Atomic mass: Ca = 40, H = 1, O = 16)

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Q10. Law of Conservation of Mass Based Question

If 15 g of Hydrogen reacts with 120 g of Oxygen, what will be the total mass of products formed?

✅ Solutions to Numerical Questions

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Q1. Balance the equation

$$H_2 + O_2 \rightarrow H_2O$$H2​+O2​→H2​O

Step 1: Count atoms

LHS: H = 2, O = 2
RHS: H = 2, O = 1

Step 2: Balance Oxygen

Put 2 before H₂O:$$H_2 + O_2 \rightarrow 2H_2O$$H2​+O2​→2H2​O

Now RHS: H = 4, O = 2

Step 3: Balance Hydrogen

Put 2 before H₂:$$\boxed{2H_2 + O_2 \rightarrow 2H_2O}$$2H2​+O2​→2H2​O​

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Q2. Balance the equation

$$Fe + O_2 \rightarrow Fe_2O_3$$Fe+O2​→Fe2​O3​

Step 1: Balance Fe

Put 2 before Fe:$$2Fe + O_2 \rightarrow Fe_2O_3$$2Fe+O2​→Fe2​O3​

Step 2: Balance Oxygen

Fe₂O₃ has 3 O, O₂ has 2 O → LCM of 2 and 3 is 6$$4Fe + 3O_2 \rightarrow 2Fe_2O_3$$4Fe+3O2​→2Fe2​O3​

✔ Balanced equation:$$\boxed{4Fe + 3O_2 \rightarrow 2Fe_2O_3}$$4Fe+3O2​→2Fe2​O3​​

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Q3. Balance the equation

$$Al + O_2 \rightarrow Al_2O_3$$Al+O2​→Al2​O3​

LCM of O = 2 and 3 → 6$$4Al + 3O_2 \rightarrow 2Al_2O_3$$4Al+3O2​→2Al2​O3​

✔ Balanced equation:$$\boxed{4Al + 3O_2 \rightarrow 2Al_2O_3}$$4Al+3O2​→2Al2​O3​​

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Q4. 4 g of H₂ reacts — Find O₂ required

Balanced equation:$$2H_2 + O_2 \rightarrow 2H_2O$$2H2​+O2​→2H2​O

Step 1: Molar masses

H₂ = 2 g
O₂ = 32 g

Step 2: From equation

2H₂ (4 g) reacts with O₂ (32 g)

So, 4 g H₂ requires 32 g O₂$$\boxed{Answer = 32\text{ g of Oxygen}}$$Answer=32 g of Oxygen​

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Q5. 32 g O₂ reacts — Find water formed

From equation:
32 g O₂ → 2H₂O

2H₂O = 2 × 18 = 36 g$$\boxed{Answer = 36\text{ g of Water}}$$Answer=36 g of Water​

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Q6. 12 g Carbon burns — Find CO₂ formed

Reaction:$$C + O_2 \rightarrow CO_2$$C+O2​→CO2​

Molar mass:
C = 12 g
CO₂ = 44 g

So,
12 g C → 44 g CO₂$$\boxed{Answer = 44\text{ g of CO}_2}$$Answer=44 g of CO2​​

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Q7. 56 g Fe reacts — Find Fe₂O₃ formed

Balanced equation:$$4Fe + 3O_2 \rightarrow 2Fe_2O_3$$4Fe+3O2​→2Fe2​O3​

Molar mass:
4Fe = 4 × 56 = 224 g
2Fe₂O₃ = 2 × 160 = 320 g

So,
224 g Fe → 320 g Fe₂O₃

Given 56 g Fe:$$\frac{320}{224} \times 56 = 80 \text{ g}$$224320​×56=80 g $$\boxed{Answer = 80\text{ g Fe}_2O_3}$$Answer=80 g Fe2​O3​​

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Q8. 24 g Mg reacts — Find MgO formed

Reaction:$$2Mg + O_2 \rightarrow 2MgO$$2Mg+O2​→2MgO

2Mg = 48 g
2MgO = 80 g

So,
48 g Mg → 80 g MgO

24 g Mg → ?$$\frac{80}{48} \times 24 = 40\text{ g}$$4880​×24=40 g $$\boxed{Answer = 40\text{ g MgO}}$$Answer=40 g MgO​

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Q9. 10 g Ca reacts — Find H₂ formed

Reaction:$$Ca + 2H_2O \rightarrow Ca(OH)_2 + H_2$$Ca+2H2​O→Ca(OH)2​+H2​

Molar mass:
Ca = 40 g
H₂ = 2 g

So,
40 g Ca → 2 g H₂

10 g Ca → ?$$\frac{2}{40} \times 10 = 0.5\text{ g}$$402​×10=0.5 g $$\boxed{Answer = 0.5\text{ g Hydrogen}}$$Answer=0.5 g Hydrogen​

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Q10. Law of Conservation of Mass

Given:
15 g H + 120 g O

Total mass of reactants:$$15 + 120 = 135\text{ g}$$15+120=135 g

According to Law of Conservation of Mass:$$\boxed{Mass\ of\ products = 135\text{ g}}$$Mass of products=135 g​

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📌 Final Quick Formula for Numericals

$$\text{Required mass} = \frac{\text{Given mass} \times \text{Required molar mass}}{\text{Given molar mass}}$$Required mass=Given molar massGiven mass×Required molar mass​