Question

Iron (III) oxide (rust) sits in water for such a long period of time that it disappears. If 2.50 moles of rust is added to excess water, how many moles of product will form?

Answer 1

Answer:

5 moles

Explanations:

The chemical reaction between Iron(III) oxide and water is expressed as:

$Fe_2O_3+3H_2O\rightarrow2Fe(OH)_3$

Given the following parameters

Moles of Iron(III)oxide = 2.5moles

According to stoichiometry, 1 mole of Iron(III)oxide produces 2 moles of Iron hydroxide, the moles of Iron hydroxide produced will be:

$\begin{gathered} moles\text{ of Iron hydroxide}=2\times2.50 \\ moles\text{ of Iron hydroxide}=5.0moles \end{gathered}$

Hence the moles of product formed is 5 moles