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Calculate the Mass of Iron(III) oxide (rust) produced by the reaction of 500.0g of iron with oxygen from the air.

User Quasdunk
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Step-by-step explanation:

We have a reaction between iron and oxygen from the air and Iron (III) oxide is produced.

First we have to determine the formulas of our compounds. Iron is a solid and its formula is just Fe. Oxygen is a diatomic gas and its formula is O₂. The formula of iron (III) oxide is Fe₂O₃. Why? Because the oxidation state (or charge) of oxygen in oxides is always 2- and the oxidation state or charge or iron in this case is 3+ (because between brackets we have the number III). Oxides are neutral compounds, if we want to form a neutral compound we will have to:

So the unbalanced reaction is:

___ Fe + ___ O₂ ---> ____ Fe₂O₃

The first step to balance an equation is to determine the number of atoms of each element that we have on both sides.

___ Fe + ___ O₂ ---> ____ Fe₂O₃

Fe: 1 Fe: 2

O: 2 O: 3

We have one atom of Fe on the left side of the equation and 2 atoms of Fe on the right side, so we have to change the coefficient for Fe on the left and write a 2.

2 Fe + ___ O₂ ---> ____ Fe₂O₃

Fe: 2 Fe: 2

O: 2 O: 3

Now Fe is balanced, but O is still unbalanced. We have 3 atoms of O on the right side of the equation and just 2 on the left side. The coefficient for O₂ that balances the equation should be 3/2.

2 Fe + 3/2 O₂ ---> ____ Fe₂O₃

Fe: 2 Fe: 2

O: 3 O: 3

But we can't have fractions as coefficients, they must be entire numbers. So we will have to multiply all the coefficients by 2 to get rid off the fraction.

4 Fe + 3 O₂ ---> 2 Fe₂O₃

Fe: 4 Fe: 4

O: 6 O: 6

Now the equation is balanced.

First, to determine the mass of iron (III) oxide we will have to convert 500.0 g of Fe into moles. To do that we need the molar mass of Fe. Then using the molar ratio between Fe and Fe₂O₃ we can find the number of moles produced. And finally with the molar mass of Fe₂O₃ we can find the mass of Fe₂O₃.

molar mass of Fe = 55.85 g/mol

molar mass of Fe₂O₃ = 159.69 g/mol

mass of Fe₂O₃ = 500.0 g of Fe * 1 mol of Fe/(55.85 g/mol) * 2 moles of Fe₂O₃/(4 moles of Fe) * 159.69 g/mol

mass of Fe₂O₃ = 714.8 g

Answer: the mass of iron (III) oxide produced is 714.8 g

Calculate the Mass of Iron(III) oxide (rust) produced by the reaction of 500.0g of-example-1
User Reginald Blue
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