Question

One of the main components of rust is a hydrated iron (III) oxide – Fe2O3•nH2O. This compound can react with acids such as hydrochloric acid. The unbalanced reaction is shown below: Fe2O3•nH2O(s) + HCl(aq)  FeCl3(aq) + H2O(l) If it takes 103 mL of 1.25 M HCl to completely react with 5.00 grams of Fe2O3•nH2O then what is the value of n? You must show all of your work to receive any credit. Round your value of n to the nearest whole number.

Answer 1

Moles of HCl reacted = molarity x volume

= 1.25 mol/L x 103 mL x 1L/1000 mL

= 0.12875 mol

The unbalanced reaction

Fe2O3•nH2O(s) + HCl(aq) = FeCl3(aq) + H2O(l)

The balanced reaction without nH2O

Fe2O3 + 6HCl(aq) = 2FeCl3(aq) + 3H2O(l)

Moles of Fe2O3 required = 0.12875/6 = 0.021458 mol

Mass of Fe2O3 = moles x molecular weight

= 0.021458 mol x 159.69 g/mol

= 3.3623 g

Mass of nH2O = mass of Fe2O3•nH2O - mass of Fe2O3

= 5.00 - 3.3623

= 1.637 g

Moles of nH2O = mass/molecular weight

= (1.637 g ) / (18 g/mol)

= 0.091 mol

Ratio of Fe2O3 : H2O = 0.021458 : 0.091

Divide by 0.021458

= 1 : 4.2

n = 4

The balanced reaction

Fe2O3•4H2O(s) + 6HCl(aq) = 2FeCl3(aq) + 7H2O(l)