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What mass of aluminum (in g) would be required to completely react with 1.00 L of 0.450 M HBr in the following chemical reaction?
2 Al(s) + 6 HBr(aq) → 2 AlBr₃ (aq) + 3 H₂(g)

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Answer:

4.0473g. Attached in image are the calculations I used to reach this answer.

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