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Chemical Reactions - Word and Chemical Equations

Grade 9IB

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

🔑Concepts

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A chemical reaction is a process where substances (reactants) transform into new substances (products) through the breaking and forming of chemical bonds.

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A Word Equation uses the names of the substances to describe the reaction, such as Iron+Oxygen→Iron(III)oxideIron + Oxygen \rightarrow Iron(III) oxide.

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A Chemical Equation uses chemical symbols and formulae to represent the reaction, for example: 4Fe(s)+3O2(g)→2Fe2O3(s)4Fe(s) + 3O_2(g) \rightarrow 2Fe_2O_3(s).

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The Law of Conservation of Mass states that mass is neither created nor destroyed in a chemical reaction. Therefore, the number of atoms of each element must be equal on both sides of the equation.

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State Symbols are used to indicate the physical state of the substances: (s)(s) for solid, (l)(l) for liquid, (g)(g) for gas, and (aq)(aq) for aqueous (dissolved in water).

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Balancing Equations involves adding coefficients (numbers in front of formulas) to ensure the number of atoms is conserved. Subscripts within a formula (like the 22 in H2OH_2O) must never be changed.

📐Formulae

Reactants→ProductsReactants \rightarrow Products

Total Mass of Reactants=Total Mass of Products\text{Total Mass of Reactants} = \text{Total Mass of Products}

2H2(g)+O2(g)→2H2O(l)2H_2(g) + O_2(g) \rightarrow 2H_2O(l)

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

💡Examples

Problem 1:

Write the balanced chemical equation for the combustion of Magnesium ribbon in Oxygen gas to form Magnesium Oxide powder.

Solution:

2Mg(s)+O2(g)→2MgO(s)2Mg(s) + O_2(g) \rightarrow 2MgO(s)

Explanation:

Start with the word equation: Magnesium+Oxygen→MagnesiumOxideMagnesium + Oxygen \rightarrow Magnesium Oxide. Write the formulas: Mg+O2→MgOMg + O_2 \rightarrow MgO. To balance the oxygen atoms, add a coefficient of 22 in front of MgOMgO, giving Mg+O2→2MgOMg + O_2 \rightarrow 2MgO. Finally, balance the magnesium atoms by adding a 22 in front of MgMg on the reactant side.

Problem 2:

Balance the following skeleton equation: Al(s)+H2SO4(aq)→Al2(SO4)3(aq)+H2(g)Al(s) + H_2SO_4(aq) \rightarrow Al_2(SO_4)_3(aq) + H_2(g)

Solution:

2Al(s)+3H2SO4(aq)→Al2(SO4)3(aq)+3H2(g)2Al(s) + 3H_2SO_4(aq) \rightarrow Al_2(SO_4)_3(aq) + 3H_2(g)

Explanation:

  1. There are 22 Aluminum atoms on the right, so we place a 22 before AlAl. 2. There are 33 sulfate (SO42−)(SO_4^{2-}) groups on the right, so we place a 33 before H2SO4H_2SO_4. 3. This gives 3×2=63 \times 2 = 6 Hydrogen atoms on the left, so we place a 33 before H2H_2 on the right to get 3×2=63 \times 2 = 6 Hydrogen atoms.