Question #b785a

1 Answer
Sep 25, 2017

#x = 3#


Your goal here is to figure out the number of moles of water of hydration that are present per mole of anhydrous iron(III) oxide.

Now, heating the sample of iron(III) oxide hydrate will cause the water of hydration to evaporate, leaving behind the anhydrous salt.

This means that you can find the mass of water of hydration present in your sample by using the Law of Mass Conservation

#overbrace("0.525 g")^(color(blue)("mass of hydrate")) = overbrace("0.400 g")^(color(blue)("mass of anhydrous salt")) + overbrace("? g")^(color(blue)("mass of water of hydration"))#

You can thus say that your sample contained

#"0.525 g " - " 0.400 g" = "0.125 g"#

of water of hydration.

Now, you need to figure out how many moles of water are present per mole of anhydrous iron(III) oxide. To do that, use the molar mass of water and the molar mass of anhydrous iron(III) oxide to convert the two masses to moles.

#0.400 color(red)(cancel(color(black)("g"))) * ("1 mole Fe"_2"O"_3)/(159.69color(red)(cancel(color(black)("g")))) = "0.002505 moles Fe"_2"O"_3#

#0.125 color(red)(cancel(color(black)("g"))) * ("1 mole H"_2"O")/(18.015color(red)(cancel(color(black)("g")))) = "0.006939 moles H"_2"O"#

To find the mole ratio that exists between anhydrous iron(III) oxide and water in the hydrate, divide both values by the smallest one

#"For Fe"_2"O"_3:" " (0.002505 color(red)(cancel(color(black)("moles"))))/(0.006939color(red)(cancel(color(black)("moles")))) = 0.361#

#"For H"_2"O": " " (0.006939 color(red)(cancel(color(black)("moles"))))/(0.006939 color(red)(cancel(color(black)("moles")))) = 1#

Notice that in order to convert this ratio to a whole number ratio, you need to multiply both values by #3#. This will get you

#"For Fe"_2"O"_3: " " 0.361 * 3 = 1.083 ~~ 1#

#"For H"_2"O": " " 1 * 3 = 3#

This means that the hydrate contains anhydrous iron(III) oxide and water of hydration in a #1:3# mole ratio, which implies that #x = 3#.

#"Fe"_2"O"_3 * x"H"_2"O" = "Fe"_2"O"_3 * 3"H"_2"O"#