If 5.3 g of gallium reactions with 5.3 g of oxygen according to the following reaction, how
many grams of gallium oxide can be produced?

Word equation:
Formula equation:
(Must be balanced!)

7.5g

Explanation:

Given parameters:

Mass of gallium = 5.3g

Mass of oxygen  = 5.3g

Unknown:

Mass of gallium oxide produced  = ?

Solution:

To solve this problem, let us write the reaction equation:

   Gallium reacts with oxygen to produce gallium oxide

                     4Ga     +   3O₂     →   2Ga₂O₃

Let us determine the limiting reactant from the equation. This reactant is the one in short supply and it determines the extent of the reaction.

        Number of moles  =

  Number of moles of Ga  =   = 0.08mole

   Number of moles of O₂ =   = 0.17mole

From the balance reaction equation;

                4 moles of Ga reacted with 3 moles of O₂

               0.08 moles of Ga will require     = 0.06 moles of O₂

But we were given 0.17 moles. This implies that oxygen gas is in excess

  If :

             4 moles of Ga gives 2 moles of Ga₂O₃

           0.08 moles of Ga will produce   = 0.04moles

Mass of Ga₂O₃ = number of moles x  molar mass

         Molar mass of Ga₂O₃ = 2(69.7)  + 3(16)  = 187.4g/mol

     Mass of Ga₂O₃ = 0.04 x 187.4 = 7.5g

hey there!

mass = 6.75 g

volume =   9 cm³

therefore:

d = mass / volume

d   = 6.75 / 9

d = 0.7 g/cm³

answer b