# Find the amount of Barium Sulphate precipitated

## Homework Statement

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 ?

## Homework 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?

## 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.

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Where did that second reaction equation come from? It's wrong.

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).

@ 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 ?

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"The solution is treated with BaCl2 solution"[/I]

Perhaps, this means that H2O is also involved ?

In that case, the second equation would be:

SO3 + HCl + BaCl2 + H2O = BaSO4 + 3HCl
Yup, that looks much better.