Aluminum sulfate is analyzed, and the sample contains 0.0670 moles of sulfate ions. how many moles of aluminum are in the sample?

0.0447 moles of Aluminum

Explanation:

Break this down to an ionic equation (which may not exist in the real world but it will give you the mol ratios).

Al2(SO4)3 ===> 2Al^(+3) + 3 SO4^(-2)

Now write the equation showing the actual number of moles of SO4^(-2) present. Indicate the Al moles with an x

Al2(SO4)3 ===> 2Al^(+3) + 3 SO4^(-2)

                                x             0.0670  

Now set  up a proportion

2/x = 3/0.0670                             Cross multiply

2*0.0670 = 3x                              Simplify the left

0.134 = 3x                                     Divide both sides by 3

0.134/3 = 3x/3                              Divide

0.0447 = x

there are 0.0447 moles of Aluminum

c. orbital shape i hope this is correct