Question

If you have 0.08157 moles of Al(ClO4)3…

a) How many moles of Al3+ ions do you have?
b) How many moles of ClO4 – ions do you have?
c) How many moles of oxygen atoms do you have?

Answer 1

Aluminum chlorate ionizes as follows,

$Al(ClO_{4})_{3} (aq) --> Al^{3+}(aq) + 3ClO_{4}^{-}(aq)$

1 mol Aluminum chlorate ionizes to give 1 mole Aluminum ion and 3 moles of chlorate ion. Each mole aluminum ion has 4 * 3 = 12 moles of Oxygen atom.

Moles of $Al(ClO_{4})_{3}$ = 0.08157 mol

a) Moles of $0.08157 mol Al(ClO_{4})_{3}} * \frac{1 mol Al^{3+}}{1 mol Al(ClO_{4})_{3}}$ = $0.08157 mol Al^{3+}$

b)Moles of $ClO_{4}^{-} = 0.08157 mol Al(ClO_{4})_{3} * \frac{ 3 mol ClO_{4}^{-}}{1 mol Al(ClO_{4}^{-}} = 0.2447 mol ClO_{4}^{-}$

c) Moles of O atom = $0.08157 mol Al(ClO_{4})_{3} *\frac{12 mol O}{1 mol Al(ClO_{4})_{3}} = 0.9788 mol O$