Molarity and Reactants

  1. Please check my work!

    1) In the preparation of a solution of sulfuric acid, 26.7g of SO3 was dissolved in enough water to prepare 8.20L of solution. What is the concentration of sulfur (moles per liter) in this solution?

    Equation given: SO3 (g) + H2O (l) --> H2SO4 (aq), which is already balanced

    M = mol of solute/ L of solution

    I don't know if this is correct because I am confused about finding the concentration of just sulfur:

    26.7 g SO3 (1 mol SO3 / 80.07 g SO3) = 0.33346 mol SO4

    Now is it (0.33326 mol)/ 8.20 L = M = 0.406659????



    2) The acid-base neutralization reaction below is often used as a demonstration of a "spontaneous" endothermic process. In this reaction solid ammonium chloride reacts with solid barium hydroxide to form barium chloride, ammonia, and water. If 8.0g of solid ammonium chloride are mixed with 62.1g of solid barium hydroxide, how many grams of solid barium chloride are formed? Report your answer to two decimal places.

    Equation given: NH4Cl (s) + Ba(OH)2 (s) --> BaCl2 (s) + NH3 + H2O (l)

    How I Balanced It: 2NH4Cl (s) + Ba(OH)2 (s) --> BaCl2 (s) + 2NH3 + 2H2O(l)

    80 g NH4Cl (1 mol NH4Cl / 53.492 g NH4Cl) = 1.49556 mol NH4Cl

    62.1 g Ba(OH)2 (1 mol Ba(OH)2 / 171.346 g Ba(OH)2) = 0.36242 mol Ba(OH)2

    Now the limiting reactant is Ba(OH)2 since there is excess NH4Cl ???

    0.36242 mol Ba(OH)2 ( 1 mol BaCl2 / 1 mol Ba(OH)2 ) (208.23 g BaCl2 / 1 mol BaCl2) = 75.47 g BaCl2 ?????



    3) In the preparation of a solution of hydrochloric acid, 13.200g of hydrogen chloride was dissolved in enough water to prepare 5.900L of solution. What is the molarity of this solution?

    13.200 g HCl (1 mol / 36.458 g HCl) = 0.36206 mol HCl

    0.36206 mol HCl / 5.900 L = 0.06137 M ????

    THANK YOU.
     
  2. jcsd
  3. Borek

    Staff: Mentor

    1. OK (although you abuse significant digits)

    2. 8.0g or 80g of NH4Cl?

    3. OK
     
  4. 1. 0.407 M, then?

    2. Sorry! It's 8.0 g for NH4Cl.

    Here I go again:

    8.0 g NH4Cl (1 mol NH4Cl / 53.492) = .1495551 mol NH4Cl

    Now NH4Cl is the limiting reactant because it's supposed to have twice the mol of Ba(OH)2 but it does not???

    .1495551 mol NH4Cl (1 mol BaCl2 / 2 mol NH4Cl) (208.23 g BaCl2 / 1 mol BaCl2) = 15.57 g BaCl2 ???

    Thanks.
     
    Last edited: Oct 2, 2005
  5. Any volunteers?

    Thanks.
     
  6. Will anyone confirm if I did #2 correctly with 8 g not 80 g?

    Thanks.
     
  7. Anyone at all please?

    Thanks.
     
  8. Borek

    Staff: Mentor

    OK.
     
Know someone interested in this topic? Share a link to this question via email, Google+, Twitter, or Facebook