Aluminum metal dissolves in hydrochloric acid. What volume, in mL, of 1.58 M HCl is needed to completely dissolve 3.200 g of aluminum?(Hint: write the single-replacement reaction first)
The Attempt at a Solution
I got the single-replacement reaction:
2Al(s) + 6HCl(aq) = 2AlCl(aq) + 3H(g)
But from there, I'm not really sure what to do. I think stoichiometry's involved, but I'm not sure how to use it.