Ionic reactions

Some physical processes, as the dissociation of some molecules into ions when they dissolve in water, are described by "ionic reactions"

When dissolving sodium chloride in water, we describe it by the following ionic reaction:-

Nacl(s) ------------ Na+ (aq) + Cl – (aq)

The previous reaction states that a solid mole of NaCl produces a mole of positive sodium ions (6.02 x1023 ions) and a mole of negative chlorine ions (6.02 x1023 ions) when it dissolves in water

When sulphuric acid reacts with sodium hydroxide forming sodium sulphate and water (Neutralization reaction) , we describe the reaction as the following:-

2NaOH(aq) + H2SO4(aq)------ Na2SO4(aq) + 2H2O (l)

In neutralization reactions, we don't need to write the signs of ions in their ionic reactions.

When adding potassium chromate (K2C2O7) to silver nitrate solution (AgNO3), insoluble silver chromate(Ag2Cr2O 7) is formed as a red ppt.

K2Cr2O7(aq) + 2AgNO3(aq) --------- 2KNO3(aq) + Ag2Cr2O7

Example (1)

Calculate the no. of carbon molecules in 48 grams of carbon (C=12)

Solution:-

The mole of carbon = 12+12 = 24g (diatomic element)

Therefore, the no. of moles = 48 /24 = 2 moles

The no. of molecules = 2 x 6.02x1023 = 12.04x1023 atoms

Example(2)

Calculate the no. of carbon atoms in 50 grams of calcium carbonate CaCO3

(Ca = 40, C=12, O=16)

Solution:-

One mole of CaCO3 = 40 + 12 + (16x3) = 100 gm

Therefore, 100 gm (one mole) of CaCO3 contains one mole of carbon

50 gm of CaCO3 = 0.5 mole of CaCO3 = 0.5 mole of carbon

The no. of carbon atoms = 0.5 x 6.02x1023 = 3.01x1023 atoms.

It's known that the volume of gas is the volume of its container, but scientists discovered that moles of all gases occupy certain volume of 22.4 litres if they are put in certain conditions called "Standard temperature and pressure (STP)"

STP: The presence of matter in temperature of 0 degree Celsius (273 Kelvin) and pressure of 760 mm.Hg (1 atomic pressure)

This means that a mole of methane gas (CH4) occupies volume of 22.4 litres (if it's in (STP), and the same to a mole of Hydrogen gas (H2) and any gas

Volume of gas (in litres) = 22.4 x no. of moles

Example (1)

Calculate the volume of 64 gm of oxygen gas in STP conditions (O=16)

Solution:-

If one mole of oxygen =16+16= 32 gm (diatomic element)

The no. of moles = 64 /32= 2 moles

The volume of oxygen gas = 22.4 x the no. of moles = 22.4 x2 =44.8L

Example:-

Calculate the volume of oxygen required for 90 g of water when reacting with hydrogen in STP (O= 16, H=1)

Solution:-

2H 2(g) + O2(g)------- 2H2O

2 mol 1 mol 2 mol

One mole of water = 16 + 1x2 = 18 grams

If one mole of oxygen produces 2 moles of water(36 grams of water)

Therefore, The no. of moles in oxygen = 90/36 = 2.5 mol

The volume of oxygen = 22.4 x2.5 = 50 L