The process for a piece of iron corroding to form rust, Fe₂O₃* H₂O, is given in the two reactions below. How much rust will be formed if 25.0 g of Fe (molar mass = 55.84) completely corrodes to form Fe₂O₃ *H₂O (molar mass = 177.70) ?
4 Fe + 12 H⁺ + 3 O₂ $\rightarrow$ 4 Fe³⁺ + 6 H₂O
2 Fe³⁺ + 4 H₂O $\rightarrow$ Fe₂O₃ *H₂O + 6 H⁺

Question

The process for a piece of iron corroding to form rust, Fe₂O₃* H₂O, is given in the two reactions below. How much rust will be formed if 25.0 g of Fe (molar mass = 55.84) completely corrodes to form Fe₂O₃ *H₂O (molar mass = 177.70) ?
4 Fe + 12 H⁺ + 3 O₂

\rightarrow

4 Fe³⁺ + 6 H₂O
2 Fe³⁺ + 4 H₂O

\rightarrow

Fe₂O₃ *H₂O + 6 H⁺

Quizplus · Accepted Answer

First, calculate the moles of Fe used: 25.0 g / 55.84 g/mol = 0.447 moles. From the balanced equations, 4 moles of Fe produce 1 mole of Fe_2O_3*H_2O. Therefore, 0.447 moles of Fe will produce 0.447/4 = 0.11175 moles of Fe_2O_3*H_2O. The mass of rust formed is then 0.11175 moles * 177.70 g/mol = 19.85 g, which rounds to 19.9 g, but due to the stoichiometry of Fe to rust being 1:1 in the overall reaction, the correct calculation should consider the total conversion of Fe to rust, leading to a doubling of the initial calculation, hence 39.8 g.

The Process for a Piece of Iron Corroding to Form →\rightarrow→

The Process for a Piece of Iron Corroding to Form $\rightarrow$