Final answer:
The concentration of Al in Al2(SO4)3 0.15 M solution is 0.30 M since each mole of Al2(SO4)3 produces 2 moles of Al3+ ions.
Step-by-step explanation:
To calculate the concentration of Al in Al2(SO4)3 0.15 M, we need to consider the stoichiometry of the compound. The formula indicates each mole of aluminum sulfate produces 2 moles of aluminum ions. Therefore, you multiply the concentration of the aluminum sulfate by 2 to find the concentration of aluminum ions:
[Al3+] = 2 × 0.15 M = 0.30 M.
The correct answer is d. 0.30 M, which represents the final concentration of aluminum ions (Al3+) in solution.