1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Find the amount of Barium Sulphate precipitated

  1. Jul 30, 2011 #1
    1. The problem statement, all variables and given/known data

    8 grams of Sulfur is burnt in Oxygen to form SO2 which is oxidized by Chlorine water. The solution is treated with BaCl2 solution. The amount of BaSO4 precipitated is ?

    2. Relevant equations

    SO2 + Cl2 + H2O = SO3 + 2HCl

    SO3 + HCl + BaCl2 = BaSO4 + HCl

    I do not know if the second one is right and if right, how to balance it?

    3. The attempt at a solution

    Prima facie it appears that 0.25 mol of S reacts with 0.25 mol of O2 to produce 0.25 mol of SO2.

    Beyond this, I do not know how to proceed since the second equation is not balanced.
  2. jcsd
  3. Jul 30, 2011 #2
    Where did that second reaction equation come from? It's wrong.
  4. Jul 30, 2011 #3
    yea if they are having you make the equations the chlorine water is a little confusing. Do they mean aqueous hydrochloric acid? You also need to use the equation for the burning of sulfur in oxygen (a reaction I do not recommend trying to do).
  5. Jul 30, 2011 #4
    @ zaldar:

    The first equation should be S + O2 = SO2.

    Since Sulfur is only 8 grams (one-fourth of a mole), the quantities involved in this equation are all 0.25 mole. That is easy to figure out.

    The question is correct. The answers are

    1. 1 mole
    2. 0.5 mole
    3. 0.4 mole
    4. 0.25 mole

    I know it is only a question of using the law of equivalent proportions, but I am unable to find the right equations.

    "The solution is treated with BaCl2 solution"

    Perhaps, this means that H2O is also involved ?

    In that case, the second equation would be:

    SO3 + HCl + BaCl2 + H2O = BaSO4 + 3HCl

    Since we originally started off with 0.25 mole of S & O2 and now we have an equivalent proportion of Ba i.e. 0.25 mole, I guess the answer is 0.25 mole. Am I right ?
    Last edited: Jul 30, 2011
  6. Jul 31, 2011 #5
    Yup, that looks much better.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook