In a solution of Al2(SO4)3 the Al3+ concentration is .12 M. What mass of Al2(SO4)3 is in 50 mL of this solution. Please explain the steps. Thanks. The Al3+ part confuses me the most.

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  1. If Al^3+ is 0.12M, that means you must have 0.06M Al2(SO4)3. If you want to do it mathematically, then
    0.12 M Al^3+ x (1 mole Al2(SO4)3/2 moles Al^3+) = 0.06 M.
    Then we know 0.06M means 0.06 moles/L or 0.06 moles/1000 mL. The amount in 50 mL must be
    0.06 moles x (50/1000) =?? moles.

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  2. But since Al3+ is only one Al doesn't it equal the moles of the compound, and Al2 would be .24 M? Basically why is Al3+ considered Al2? Isnt Al3 just the ionic version of Al?

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  3. In a solution of Al2(SO4)3, the Al3+ concentration is 0.12 M. What mass of Al2(SO4)3 is in 50 mL of this solution?
    Who can solve please..please help me

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  4. In a solution of Al2(SO4)3, the Al3+ concentration is 0.12 M. What mass of Al2(SO4)3 is in 50 mL of this solution?
    email me jgbunsay for the answer. I can't pay online tutor ..please help

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