Ask questions and get helpful answers.

Suppose we collect hydrogen gas, H2(g), over water at 21◦C in a vessel with total pressure of 743 Torr. If the hydrogen gas is produced by the reaction of aluminum with hydrochloric acid:

2Al(s) + 6HCl(aq) → 2AlCl3(aq) + 3H2(g)
what volume of hydrogen gas will be collected if 1.35 g Al(s) reacts with excess HCl(aq)? Express
your answer in liters.

  1. 👍
  2. 👎
  3. 👁
  4. ℹ️
  5. 🚩

2 answers

  1. 2Al + 6HCl => 2AlCl3 + 3H2
    (1.35/27)mole Al = 0.05molAl => 3/2(.05)mol H2 = 0.075mol H2
    Vol H2 = nRT/P = [(0.075)(0.08206)(294)(760)]/(743) = 1.85L H2

    1. 👍
    2. 👎
    3. ℹ️
    4. 🚩
  2. 1.9L

    1. 👍
    2. 👎
    3. ℹ️
    4. 🚩

Answer this Question

Related Questions

Still need help?

You can ask a new question or browse existing questions.