For the reaction 2Fe+o2 -->Feo how many grams of iron(ll) oxide are produced from 479.6 grams of iron in an excess of oxygen (Fe=56gmol, O=16g mol)

Mass of iron(ll) oxide= 616.608 g

Given

Reaction

2Fe+O2 -->2FeO

479.6 grams of iron

Required

mass of iron(ll) oxide

Solution

mol of iron :

= mass : Ar Fe

= 479.6 g : 56 g/mol

= 8.564

From the equation, mol FeO :

= 2/2 x mol Fe

= 2/2 x 8.564

= 8.564 moles

Mass of iron(ll) oxide :

= mol x MW

= 8.564 x 72 g/mol

= 616.608 g

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