Question

First second and third ionization energies of aluminum are 578 ; 1817;2745 calculated energy required to convert all atoms of al to al3+ in 270 mg of al vapours

Answer 1

The equations representing the successive ionization of Al to $Al^{3+}$,

$Al(g) --> Al^{+} + e^{-} IE_{1} = 578 kJ/mol$

$Al^{+}(g) --> Al^{2+}(g) + e^{-} IE_{2} = 1817 kJ/mol$

$Al^{2+}(g) -->Al^{3+}(g) + e^{-} IE_{3} = 2745 kJ/mol$

The net equation representing the ionization of Al to $Al^{3+}$ can be obtained by adding all the three above equations.

$Al(g)--> Al^{3+}(g) + 3e^{-} IE = IE_{1}+IE_{2}+IE_{3}$

IE = (578 + 1817 + 2745) kJ/mol = 5140 kJ/mol

Given mass of Al = 270 mg

Converting mass to moles:

$270 mg * \frac{1 g}{1000 mg} * \frac{1 mol Al }{27 g Al} = 0.01 mol Al$

Calculating the energy required to convert 270 mg Al to $Al^{3+}$:

$0.01 mol Al * \frac{5140 kJ}{1 mol Al} =51.4 kJ$

Therefore, 51.4 kJ of energy is required to convert 270 mg Al to $Al^{3+}$

Answer 2

Answer:

that first we have to all three ionization then we have to find mass to mol