What mass of aluminum is produced by the decomposition of 5.0 kg A l 2O 3? How many moles of oxygen are produced?

mass of Aluminum produced = = 2.647 kg

moles of Oxygen = 73.53 moles

Given

5.0 kg Al₂O₃

Required

moles of Oxygen

Solution

The aluminum oxide gets decompose into aluminum and oxygen gas.

Reaction :

2Al₂O₃(s)⇒ 4 Al(s) + 3O₂(g)

mol of Al₂O₃(MW=102 g/mol) :

= mass : MW

= 5000 g : 102 g/mol

= 49.02 moles

From the equation, mol ratio of Al₂O₃ : O₂ = 2 : 3, sol mol O₂ :

= 3/2 x mol Al₂O₃

= 3/2 x 49.02

= 73.53 moles

mol Al :

= 4/2 x mol Al₂O₃

= 2 x 49.02

= 98.04 moles

Mass Al :

= mol x Ar Al

= 98.04 x 27 g/mol

= 2647.08 g

= 2.647 kg

the grams   of   barium that   are in   2.3 x 10^23   atoms of barium   is   52.4 g (answer c)

    calculation

mass = moles   of ba   x molar   mass of ba

  molar mass of ba from periodic table=   137.3 g/mol

  use the   avogadro's law   constant   to   calculate the moles of barium

  that   is     1 mole =   6.02   x 10^23 atoms

                    ? moles = 2.3 x 10^23   atoms

by cross   multiplication

=(1 mole   x 2.3x 10^23 atoms)   / 6.02   x10^23 atoms)=   0.382   moles

mass   is therefore = 137.3   g/mol x 0.382 mol = 52.4 ( answer c)

explanation: