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)} * (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) * ( 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) *(12 mol O)/(1 mol Al(ClO_(4))_(3)) = 0.9788 mol O`