Question

Aluminum metal dissolved in hydrochloric acid as follows: 2Al(s) + 6HCl(aq) ? 2AlCl3(aq) + 3H2(g) mL is the minimum volume of 6.0 M HCl(aq) needed to completely dissolve 3.20 g of aluminum (26.98 g/mol) in this reaction. grams of AlCl3 (133.33 g/mol) would be produced by complete reaction of 3.20 g of aluminum.

          Aluminum metal dissolved in hydrochloric acid as follows:

2Al(s) + 6HCl(aq) ? 2AlCl3(aq) + 3H2(g)

mL is the minimum volume of 6.0 M HCl(aq) needed to completely dissolve 3.20 g of aluminum (26.98 g/mol) in this reaction.

grams of AlCl3 (133.33 g/mol) would be produced by complete reaction of 3.20 g of aluminum.

Aluminum metal dissolved in hydrochloric acid as follows:

2Al(s) + 6HCl(aq) ? 2AlCl3(aq) + 3H2(g)

mL is the minimum volume of 6.0 M HCl(aq) needed to completely dissolve 3.20 g of aluminum (26.98 g/mol) in this reaction.

grams of AlCl3 (133.33 g/mol) would be produced by complete reaction of 3.20 g of aluminum.

Added by Brian D.

Chemistry: Structure and Properties

Nivaldo Tro 2nd Edition

Instant Answer

Solved by Expert Shaiju T

04/13/2022

Step 1

Number of moles of Al = mass / molar mass = 3.20 g / 26.98 g/mol = 0.1186 mol Show more…

Show all steps

Thanks for your feedback!

Aluminum metal dissolved in hydrochloric acid as follows: 2Al(s) + 6HCl(aq) → 2AlCl3(aq) + 3H2(g) mL is the minimum volume of 6.0 M HCl(aq) needed to completely dissolve 3.20 g of aluminum (26.98 g/mol) in this reaction. grams of AlCl3 (133.33 g/mol) would be produced by complete reaction of 3.20 g of aluminum.