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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.
     
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