g For the following reaction, 20.9 grams of iron are allowed to react with 9.19 grams of oxygen gas . iron(s) oxygen(g) iron(II) oxide(s) What is the maximum mass of iron(II) oxide that can be formed

26.87g

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

2Fe + O2 —> 2FeO

Next, we shall determine the masses of Fe and O2 that reacted and the mass of FeO produced from the balanced equation.

This is illustrated below:

Molar mass of Fe =56 g/mol

Mass of Fe from the balanced equation = 2 x 56 = 112 g

Molar mass of O2 = 16x2 = 32 g/mol

Mass of O2 from the balanced equation = 1 x 32 = 32 g

Molar mass of FeO = 56 + 16 =72 g/mol

Mass of FeO from the balanced equation = 2 x 72 = 144 g

From the balanced equation above,

112 g of Fe reacted with 32 g of O2 to produce 144 g of FeO.

Next, we shall determine the limiting reactant.

This is illustrated below:

From the balanced equation above,

112 g of Fe reacted with 32 g of O2.

Therefore, 20.9 g of Fe will react with = (20.9 x 32)/112 = 5.97 g of O2.

From the calculations made above, we can see that only 5.97 g out of 9.19 g of O2 given were required to react completely with 20.9 g of Fe.

Therefore, Fe is the limiting reactant and O2 is the excess reactant.

Finally, we shall determine the mass of FeO produced from the reaction.

In this case, the limiting reactant will be used, as it will give the maximum yield of the reaction since all of it is used up in the reaction.

The limiting reactant is Fe and the maximum mass of FeO produced can be obtained as follow:

From the balanced equation above,

112 g of Fe reacted to produce 144 g of FeO.

Therefore, 20.9 g of Fe will react to produce = (20.9 x 144)/112 = 26.87g of FeO.

Therefore, the maximum mass of iron(II) oxide, FeO produced is 26.87g.

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