2Al(s)+ 6HCl(aq)→ 2AlCl₃(aq)+ 3H₂(g) According to the equation above,how many grams of aluminum are needed to completely react with 4.32 mol of hydrochloric acid? A)350 g B)52.6 g C)38.9 g D)4.32 g E)116.6 g

Question

Quizplus · Accepted Answer

The balanced equation shows that 2 moles of Al react with 6 moles of HCl. Therefore, 1 mole of Al reacts with 3 moles of HCl. For 4.32 moles of HCl, you need 4.32/3 = 1.44 moles of Al. The molar mass of Al is approximately 26.98 g/mol, so 1.44 moles of Al is 1.44 * 26.98 = 38.9 g.

2Al(s)+ 6HCl(aq)→ 2AlCl3(aq)+ 3H2(g)

2Al(s)+ 6HCl(aq)→ 2AlCl₃(aq)+ 3H₂(g)