Use the data in the Successive Ionization Energies and Electron Affinity tables to determine the following. (Assume the values in the Successive Ionization Energies table are given to the ones place.) (a) the electron affinity of Ar2+ kJ/mol (b) the electron affinity of Si + kJ/mol (c) the ionization energy of Cl − kJ/mol (d) the ionization energy of Cl kJ/mol (e) the electron affinity of Cl + kJ/mol

Use the data in the Successive Ionization Energies and Electron Affinity tables to determine the following. (Assume the values in the Successive Ionization Energies table are given to the ones place.)

(a)

the electron affinity of Ar2+

kJ/mol

(b)

the electron affinity of S+

kJ/mol

(c)

the ionization energy of Cl−

kJ/mol

(d)

the ionization energy of Cl

kJ/mol

(e)

the electron affinity of Cl+

kJ/mol

Successive Ionization Energies in Kilojoules per Mole for the Elements in Period 3 General increase 13 14 Element 11 12 I. 16

Explanation:

given:

mass of na = 115 g

excess cl2

to determine:

mass of nacl produced

explanation:

given reaction is-

2na(s) + cl2(g) → 2nacl(s)

since cl2 is in excess, na will be the limiting reagent

as per the reaction stoichiometry na: nacl = 1: 1

i.e. moles of na reacted = moles of nacl formed

now, # moles of na = mass of na/atomic mass

                                  = 115 g/23 g.mol-1 = 5 moles

therefore, moles of nacl = 5

molar mass of nacl = 58 g/mol

mass of nacl = 5 moles * 58 g.mol-1 = 290 g

ans: amount of nacl produced = 290 g