Answer:
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
