Answer to Question #109082 in General Chemistry for Jamie Strong

Question

Task

How many grams of iron oxide can be produced with 56.3 grams of o2 and 57.8 grams of fe

Basic explanations needed

Answer 1

Firstly, we need to check if there is one of the reagents (O₂ or Fe) in excess (it means that some amount either of O₂ or Fe might be not reacted). In order to do this we will calculate (according to chemical equation, see below) how many grams of O₂ is needed to oxidize 57.8 grams of Fe:

4Fe + 3O₂ = 2Fe₂O₃ (1)

m(O₂) = (3*31.998 g/mol * 57.8 g)/4*55.847 g/mol = 24.84 g.

31.998 g/mol - Molar mass of Oxygen (calculated from Periodic table)

55.847 g/mol - Molar mass of Iron (Fe) (calculated from Periodic table)

Therefore to oxidize 57.8 g of Fe we need only 24.84 g of O₂, what means that we have excess of O₂ and 56.3 g - 24.84 g = 31.46 g of O₂ will not react.

Secondly, we will calculate how many grams of Iron oxide (Iron (III) oxide) will form acording to reaction (1). So,

m(Fe₂O₃) = (2*159.69 g/mol*57.8 g)/4*55.847 g/mol = 82.64 g (of Fe₂O₃, iron oxide is formed)

159.69 g/mol - Molar mass of Fe₂O₃.

This amount also can be calculated using reacted mass of O₂ (24.84 g):

m(Fe2O3) = (2*159.69 g/mol*24.84 g)/3*31.998 g/mol = 82.64 g (of Fe2O3, iron oxide is formed).

Answer: 82.64 g of iron oxide can be produced. (However, some amount of O₂ (31.46 g) will not react)