2Al(s)+ 6HCl(aq)→ 2AlCl 3 (aq)+ 3H 2 (g)

The Answer of 2Al(s)+ 6HCl(aq)→ 2AlCl 3 (aq)+ 3H 2 (g)

2Al(s)+ 6HCl(aq)→ 2AlCl 3 (aq)+ 3H 2 (g)

The Answer of 2Al(s)+ 6HCl(aq)→ 2AlCl 3 (aq)+ 3H 2 (g)

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2Al(s)+ 6HCl(aq)→ 2AlCl₃(aq)+ 3H₂(g)

Question 21

Multiple Choice

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

2Al(s)+ 6HCl(aq)→ 2AlCl3(aq)+ 3H2(g)

Multiple Choice

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2Al(s)+ 6HCl(aq)→ 2AlCl₃(aq)+ 3H₂(g)

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