## Sunday, 4 January 2015

### 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 $×$ relative formula mass = 0.1 $×$ (23 + 35.3) = 0.1 $×$ 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 $×$ molar volume
= 0.05 $×$ 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 $×$ 2) + (16 $×$ 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 $×$ relative atomic mass = 0.5 $×$ 56 = 28g