What volume of oxygen is needed to produce 210.4 g of aluminum oxide from aluminum? Please solve the question

Question

Please answer the question 

Answer 1

As stated by someone else the correctly balanced equation is

4 Al + 3 O₂ = 2 Al₂O₃

since the molecular weight of Al₂O₃  is 26.98 * 2 + 16.00 * 3 = 102.0

so you have produced 210.4 g / 102.0 g/mole = 2.063 moles of aluminum oxide

 

since 3 moles of Oxygen produces 2 moles of aluminum oxide the total amount of Oxygen needed is

2.063 /2 * 3 = 3.095 moles

 

converting to liters:

3.095 moles * 22.4 liters/mole (Avogadro's law) = 69.3 liters of O₂

 

 

Answer 2

   2Al + 3O  =   Al2O3

 

   So the ratio of Aluminum to Oxygen is 2 to 3.

  Let x be the number of grams per share.

 

 2X + 3X = 210.4

 

5x = 210.4

 

X = 42.08

 

So you need 84.16 grams of Aluminum and 126.24 grams of oxygen

 

The atomic weight of Aluminum is 26.982 so you need

84.16/26 = 3.12 moles of Aluminum

 

The atomic weight of oxygen is 16 so

you need 126.24/16 = 7.89 moles of oxygen