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Chemistry · Ch 1 — Some Basic Concepts of Chemistry

Reactions in Solutions

1.10.2

Reactions in Solutions

Reactions in Solutions

A majority of reactions in the laboratories are carried out in solutions. Therefore, it is important to understand how the amount of substance is expressed when it is present in a solution. The concentration of a solution — the amount of substance in a given volume — can be expressed in any of the following ways, each explained in its own part below:

  1. Mass per cent (or weight per cent, w/w %)
  2. Mole fraction
  3. Molarity
  4. Molality

Balancing a Chemical Equation

The book sets out the method in a highlighted box, which we follow here. According to the law of conservation of mass, a balanced chemical equation has the same number of atoms of each element on both sides of the equation. Many chemical equations can be balanced by trial and error. Consider first the reactions of a few metals and non-metals with oxygen to give oxides:

4 Fe (s)+3 O2 (g)→2 Fe2O3 (s)(a) balanced equation4\,Fe\,(s) + 3\,O_2\,(g) \rightarrow 2\,Fe_2O_3\,(s) \qquad \text{(a) balanced equation}

2 Mg (s)+O2 (g)→2 MgO (s)(b) balanced equation2\,Mg\,(s) + O_2\,(g) \rightarrow 2\,MgO\,(s) \qquad \text{(b) balanced equation}

P4 (s)+O2 (g)→P4O10 (s)(c) unbalanced equationP_4\,(s) + O_2\,(g) \rightarrow P_4O_{10}\,(s) \qquad \text{(c) unbalanced equation}

Equations (a) and (b) are balanced, since there are the same number of metal and oxygen atoms on each side. Equation (c), however, is not: phosphorus atoms are balanced, but the oxygen atoms are not. To balance it, place the coefficient 5 on the left of oxygen:

P4 (s)+5 O2 (g)→P4O10 (s)balanced equationP_4\,(s) + 5\,O_2\,(g) \rightarrow P_4O_{10}\,(s) \qquad \text{balanced equation}

Now let us take the combustion of propane, C3H8C_3H_8, and balance it in steps:

Step 1 — Write down the correct formulas of reactants and products. Propane and oxygen are reactants; carbon dioxide and water are products.

C3H8 (g)+O2 (g)→CO2 (g)+H2O (l)unbalanced equationC_3H_8\,(g) + O_2\,(g) \rightarrow CO_2\,(g) + H_2O\,(l) \qquad \text{unbalanced equation}

Step 2 — Balance the number of C atoms. Since 3 carbon atoms are in the reactant, three CO₂ molecules are required on the right side.

C3H8 (g)+O2 (g)→3 CO2 (g)+H2O (l)C_3H_8\,(g) + O_2\,(g) \rightarrow 3\,CO_2\,(g) + H_2O\,(l)

Step 3 — Balance the number of H atoms. On the left there are 8 hydrogen atoms; each water molecule carries two, so four molecules of water are required.

C3H8 (g)+O2 (g)→3 CO2 (g)+4 H2O (l)C_3H_8\,(g) + O_2\,(g) \rightarrow 3\,CO_2\,(g) + 4\,H_2O\,(l)

Step 4 — Balance the number of O atoms. There are 10 oxygen atoms on the right side (3 × 2 = 6 in CO₂ and 4 × 1 = 4 in water), so five O₂ molecules are needed to supply them.

C3H8 (g)+5 O2 (g)→3 CO2 (g)+4 H2O (l)C_3H_8\,(g) + 5\,O_2\,(g) \rightarrow 3\,CO_2\,(g) + 4\,H_2O\,(l)

Step 5 — Verify that the number of atoms of each element is balanced in the final equation. The equation shows three carbon atoms, eight hydrogen atoms, and 10 oxygen atoms on each side. …

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← Previous · 1.10.1Problem 1.5 — 50.0 kg of N₂ (g) and 10.0 kg of H₂ (g) are mixed to produce NH₃ (g). …Next · 1.10.2 →Problem 1.6 — A solution is prepared by adding 2 g of a substance A to 18 g of water…