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Chemical Reactions - Balancing Equations

Grade 9IB

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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The Law of Conservation of Mass states that in a closed system, mass is neither created nor destroyed. Consequently, the number of atoms of each element must be identical in both the reactants and the products.

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A chemical equation consists of reactants on the left side of the yield arrow (ightarrow ightarrow) and products on the right side.

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Coefficients (the numbers in front of a formula, e.g., the 22 in 2H2O2H_2O) are used to balance the equation by multiplying the number of atoms in the molecule that follows.

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Subscripts (the small numbers within a formula, e.g., the 22 in O2O_2) indicate the ratio of atoms in a molecule. These must never be changed when balancing, as doing so changes the identity of the substance.

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State symbols indicate the physical state of matter: (s)(s) for solid, (l)(l) for liquid, (g)(g) for gas, and (aq)(aq) for aqueous (dissolved in water).

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Balanced equations must reflect the smallest whole-number ratio of coefficients.

📐Formulae

∑Massreactants=∑Massproducts\sum \text{Mass}_{\text{reactants}} = \sum \text{Mass}_{\text{products}}

aA+bB→cC+dDaA + bB \rightarrow cC + dD

Number of atoms=Coefficient×Subscript\text{Number of atoms} = \text{Coefficient} \times \text{Subscript}

💡Examples

Problem 1:

Balance the following combustion reaction: CH4(g)+O2(g)→CO2(g)+H2O(g)CH_4(g) + O_2(g) \rightarrow CO_2(g) + H_2O(g)

Solution:

CH4(g)+2O2(g)→CO2(g)+2H2O(g)CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(g)

Explanation:

Start by balancing Carbon: there is 11 on both sides. Next, balance Hydrogen: there are 44 on the left and 22 on the right, so we add a coefficient of 22 to H2OH_2O. Finally, count Oxygen: there are 22 on the left and 44 on the right (22 from CO2CO_2 and 22 from 2H2O2H_2O). Add a coefficient of 22 to O2O_2 to balance.

Problem 2:

Balance the synthesis of ammonia: N2(g)+H2(g)→NH3(g)N_2(g) + H_2(g) \rightarrow NH_3(g)

Solution:

N2(g)+3H2(g)→2NH3(g)N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)

Explanation:

There are 22 Nitrogen atoms on the left, so we place a 22 in front of NH3NH_3. This gives us 2×3=62 \times 3 = 6 Hydrogen atoms on the right. To match this, we place a coefficient of 33 in front of H2H_2 on the left (3×2=63 \times 2 = 6).

Problem 3:

Balance the reaction between Aluminum and Copper(II) Chloride: Al(s)+CuCl2(aq)→AlCl3(aq)+Cu(s)Al(s) + CuCl_2(aq) \rightarrow AlCl_3(aq) + Cu(s)

Solution:

2Al(s)+3CuCl2(aq)→2AlCl3(aq)+3Cu(s)2Al(s) + 3CuCl_2(aq) \rightarrow 2AlCl_3(aq) + 3Cu(s)

Explanation:

Chlorine has 22 atoms on the left and 33 on the right. The lowest common multiple is 66. We multiply CuCl2CuCl_2 by 33 and AlCl3AlCl_3 by 22. This creates a need to balance the metals: add a 22 in front of AlAl and a 33 in front of CuCu.