3.6 Chemical equations

Chemical Equations
■ A chemical equations shows :

	►	The reactant used in the reaction, written on the left.
	►	The product obtained from the reaction, written on the right of equation.
	►	The number of moles of reactants and products in the reaction.
	►	The physical state of the substances: ◉ Solid (s) ◉ Liquid (l) ◉ Gas (g) ◉ Aqueous (aq)
	►	In general: Reactant → Product
	►	Examples of chemical equations: Chemical equation in words Chemical equations Sodium hydroxide + sulphuric acid → sodium sulphate + water 2NaOH(aq) + H2SO4(aq) → Na2SO4(aq) + 2H2O(l) Copper(II) carbonate $\underset{\mathit{heating}}{\to }$copper(II) oxide + carbon dioxide ${\text{CuCO}}_3\text{(s)}\underset{\Delta }{\to }\text{CuO (s) +}{\text{CO}}_2\text{(g)}$

■ The following steps are involved when writing a chemical equation:

	►	Correctly write the chemical formulae of the reactant on the left and the products on the right of the arrow mark.
	►	Balance the equation by making sure that the number of atom of each element is the same on both sides. ○ The number of atoms of each element before and after the reaction is the same. ○ Adjust the coefficient in front of the chemical formulae and not the subscripts in the formulae.
	►	Write the physical state of each substance.

■ This video provides an explanation of balancing chemical equation.

Interpreting chemical equations qualitatively and quantitatively
■ Stoichiometry of equation:

►

The relationship between the ratios of the number of moles of reactants to the number of moles of products in a reaction..

■ Interpreting chemical equations

	►	The reactant taking part in the reaction.
	►	The products formed in the reaction.
	►	The number of moles of each substance taking part in the reaction and the number of moles of products formed.
	►	The physical states of all the reactants and products.
	►	Example: Chemical Equation Reactant Product 2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g) 2 moles of sodium sodium reacts with 2 moles of liquid water 2 moles of aqueous sodium hydroxide and 1 mole of hydrogen gas N2(g) + 3H2(g) → 2NH3(g) 1 mole of nitrogen gas reacts with 3 moles of hydrogen gas 2 moles of ammonia gas

■ This video provides explanation of solving stoichiometric equation:

Numerical problem involving chemical equations
✍ Worked-example 3.6(a)
In the reaction between sodium and chlorine given below, 2.3 g of sodium is burnt at room temperature.
2Na(s) + Cl₂(g) → 2NaCl(s)
(Molar volume at room temperature = 24dm³, Na = 23, Cl = 35.5)
Calculate:

(a)	The amount of sodium chloride produced.
	Number of moles of sodium = (Mass of Sodium / Relative atom mass) Number of moles of sodium = (2.3 / 23) Number of moles of sodium = 0.1 mol According to the equation, 2 moles of sodium produces 2 moles of sodium chloride. Thus, 0.1 mole of sodium produces 0.1 moles of sodium chloride. Mass of sodium chlorides produces =number of moles of sodium chloride $\times$ relative formula mass = 0.1 $\times$ (23 + 35.3) = 0.1 $\times$ 58.5 = 5.85g
(b)	The volume of chlorine used under room conditions in this experiment.
	According to the equation, 2 moles of sodium reacts with 1 moles of gas chlorine. Thus, 0.1 mole of sodium reacts with 0.05 moles of gas chlorine. Volume of chlorine used = number of moles of chlorine $\times$ molar volume = 0.05 $\times$ 24 = 1.2dm3

✍ Worked-example 3.6(b)
40g of iron (III) oxide is reduced by carbon powder to form iron and carbon dioxide.
(Relative atomic mass: O = 56, Fe = 56)

(a)	Write a balanced chemical equation for the reaction.
	2Fe2O3(s) + 3C(s) → 4Fe(s) + 3CO2(g)
(b)	Calculate the maximum mass of iron (III) formed.
	Number of moles of iron (III) oxide = (Mass of Iron (III) oxide / Relative formula mass) Number of moles of iron (III) oxide = 40 / [(56 $\times$ 2) + (16 $\times$ 3)] Number of moles of iron (III) oxide = 40 / 160 = 0.25mol According to the equation, 2 moles of iron (III) oxide produces 4 moles of iron. Thus, 0.25 mole of iron (III) oxide produced 0.5 moles of iron. Mass of iron produces = number of moles of iron $\times$ relative atomic mass = 0.5 $\times$ 56 = 28g

← 3.5 Chemical Formulae

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