2Al(s) + 6HCl(aq) ––––> 2AlCl3(aq) + 3H2(g) According to the equation above, how many grams of aluminum metal are needed to completely react with 3.83 mol of hydrochloric acid? A) 310 g B) 46.6 g C) 34.4 g D) 3.83 g E) 103.3 g