Simple equation of barium chloride and copper (II) sulfate

Question

An aqueous solution of barium chloride is mixed with an aqueous solution of copper (II) sulfate. If there are 5.00 grams of barium chloride in solution how many grams of barium sulfate will precipitate.

i got the equation and balanced it : BaCl2 (aq) CuSO4 (aq) >> BaSO4 (g) CuCl2 (g)

Then multiplied the 5g by 233.39g (the atomic mass of BaSO4) and then i divided that number by 208.22 (the atomic mass of BaCl2) and got the answer of 5.6 grams of BaSO4, however i was wondering if I did it right? I've seen people do it moles but that confuses me and i just did mass to mass proportion. Also, i wanted to know if this is a double replacement reaction?

Answer 1

BaCl2 (aq) + CuSO4 (aq) >> BaSO4 (s) + CuCl2 (aq)

moles BaCl2 = 5.00 g / 208.233 g/mol = 0.0240

= moles BaSO4

mass BaSO4 = 0.0240 mol x 233.393 g/mol = 5.60 g

Answer 2

I think your reaction is that of barium chloride with excess copper sulfate to form a precipitate of barium sulfate

BaCl2(aq) + CuSO4(aq) => BaSO4(s) + CuCl2 (aq) The reaction could also be written ionically.

Your reasoning is quite right

208.22 g BaCl2 yields 233.39g BASO4

1g  BaCl2 yields (233.39/208.22)g BaSO4

5g  BaCl2 yields (5 x 233.39/208.22) = 5.604g BaSO4, just as you said

And yes, it is a double replacement reaction. See:

http://web.fccj.org/~smilczan/Two5/DR.ht...

Answer 3

Yes, your answer is correct and it is a double replacement reaction.

Answer 4

This is indeed a double replacement reaction.  Assuming BaCl2 is the limiting reagent your calculations are correct.  As for the phases we know that BaSO4 should be a solid (s) as inferred by the question.