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Stoichiometry of Chemical Reaction involving Volume of Gases part two

In this second part, we are going to be looking at the stoichiometry of reaction involving the volume of gases. Just like what we saw in mass and mole, there is a relationship which also exists between mass and volume or mole and volume that is what we are going to look at. A close look at a balanced chemical equation shows the ratio in which the reactants react and the ratio in which the products are produced. However, an experiment has shown that at standard temperature and pressure, gases have an equal volume which is numerically equal to 22.4 dm 3 or 22400cm 3 . This volume is known as the molar volume of gases. 1mole of any gas at STP = 22.4dm 3 32g of oxygen at stp = 22.4dm 3 Let us consider the decomposition of potassium trioxochlorate (V) compound. 2KClO 3(s) ---------- 2KCl (s) + 3O 2(g) 2 moles of KClO 3 will decompose to give 2 moles of KCl and 3 moles of oxygen gas. Since the molar volume gas is 22.4dm 3 . The reacting mass of KClO 3 is 2x 122.

Stoichiometry of Chemical Reaction involving Mass and Mole Part one

Stoichiometry is the study of the relationships which exist between reactants and products in a chemical reaction. This relationship can easily be seen in a balanced chemical equation. For example, K 2 SO 4 ------ 2K + + SO 4 2- 1 2 1 In the ionization reaction above, notice that one-mole potassium tetraoxosulphate (VI) gives two moles of potassium ions and one mole of tetraoxosulphate (VI) ion in solution. This ionization equation also shows the mole ratio of the reactants and the products, which is known as the stoichiometry of the chemical reaction. It is necessary to note that the mole of a substance is not affected if the reactants or products are ions, atoms or molecules etc. in a chemical reaction. CaCO 3(s) + 2HCl (aq) ------ CaCl 2(s) + H 2 O (l) + CO (g) 1mole 2mole 1mole 1 mole 1mole Consider the chemical reaction above, and note the relationship between the reactants and the prod

How to apply Gay-Lussac’s Law in chemistry calculations

Gay-Lussac’s Law explains the conditions that govern the combination of gases in a chemical reaction. This law states that gases react in volumes which are in simple ratio to each other and to the volumes of the products, provided that the temperature and pressure are constant. For instance, 4NH 3(g) + 3O 2(g) = 6H 2 O (g) + 2N 2(g) In the reaction above, according to Gay-Lussac’s Law, it can be said that 4cm 3 of ammonia reacts with 3cm 3 of oxygen to give 6cm 3 of water vapour and 2cm 3 of nitrogen. This is in line with his observation in his experiment with oxygen and hydrogen and also with hydrogen and chlorine. 2H 2(g) + O 2(g) = 2H 2 O (g) 2cm 3 1cm 3 2cm 3 The ratios of the reactants and product are 2:1:2 and the moles are 2 moles, 1 mole, and 2 moles respectively. He also noticed the same pattern in the case of the reaction with hydrogen and chlorine. H 2(g) + Cl 2(g) = 2HCl (g) 1: 1: 2 1mole of hydrogen reacts with 1mole of chlo

How to prepare a standard solution of an acid

A standard solution is a solution which has a known concentration, which means the amount in mole or mass concentration of the substance which is dissolved in a known volume of a solvent is known. If you want to prepare a standard solution of an acid, the knowledge of the relationship between amount, molar concentration and volume is very necessary C = n/V Where: n = amount (mole) V = volume (dm 3 ) C = molar concentration (mole per dm 3 ) Experiment has shown that C 1 V 1 = C 2 V 2 Since the amount of solute in a solution is not affected by increase in the volume of the solvent. This formula becomes the basic principle of dilution. It is also known as dilution formula. Information required for preparation of standard solution of acid is listed below: · Specific gravity of the acid · Percentage purity of the acid · Molar mass of the acid · Concen