2.0mol of Iron(s) reacts with 48.0g of Oxygen gas to form iron(III)oxide(s). Complete the problem by choosing the correct pull down quantities in the BCA table below.

Fe and O2 = 0 mol

Fe2O3 = 1 mol

Given

2 mol Fe

48 g O

Required

Fill BCA table

Solution

mol Oxygen :

= mass : MW O2

= 48 g : 32 g/mol

= 1.5 moles

Limiting reactants :

Fe : O2 = 2/4 : 1.5/3 = 0.5 : 0.5

Both reactants have completely reacted

mol Fe2O3 :

= 2/4 x mol Fe

= 2/4 x 2

= 1

BCA table

Reaction

             4Fe + 3O2 --> 2Fe2O3

Before   2          1.5            0

Change  2         1.5             1

After       0           0              1

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