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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.

20mol Of Irons Reacts With 480g Of Oxygen Gas To Form IronIIIoxides Complete The Problem By Choosing The Correct Pull Down Quantities In The BCA Table Below class=

Sagot :

ardni313

Mol after reaction :

Fe and O2 = 0 mol

Fe2O3 = 1 mol

Further explanation

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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