Aluminum metal dissolved in hydrochloric acid as follows:
2Al(s) + 6HCl(aq) $\to$ 2AlCl₃(aq) + 3H₂(g)
a. What is the minimum volume of 6.0 M HCl(aq) needed to completely dissolve 3.20 g of aluminum in this reaction?
b. What mass of AlCl₃would be produced by complete reaction of 3.20 g of aluminum?

Question

Aluminum metal dissolved in hydrochloric acid as follows:
2Al(s) + 6HCl(aq)

\to

2AlCl₃(aq) + 3H₂(g)
a. What is the minimum volume of 6.0 M HCl(aq) needed to completely dissolve 3.20 g of aluminum in this reaction?
b. What mass of AlCl₃would be produced by complete reaction of 3.20 g of aluminum?

Quizplus · Accepted Answer

The Answer of Aluminum metal dissolved in hydrochloric acid as follows: 2Al(s) + 6HCl(aq) $\to$ 2AlCl₃(aq) + 3H₂(g) a. What is the minimum volume of 6.0 M HCl(aq) needed to completely dissolve 3.20 g of aluminum in this reaction? b. What mass of AlCl₃would be produced by complete reaction of 3.20 g of aluminum?

Aluminum Metal Dissolved in Hydrochloric Acid as Follows: 2Al(s) →\to→ 2AlCl3(aq) + 3H2(g) A

Aluminum Metal Dissolved in Hydrochloric Acid as Follows:
2Al(s) $\to$ 2AlCl₃(aq) + 3H₂(g)
A