Answer to Question #96615 in General Chemistry for Mike

Question #96615
In hope to measure SO3 quantity in oleum H2SO4*xSO3 we did experiment. To saturated Ba(OH)2 solution we poured 1 gram oleum. Reactions accured:
SO3(l)+H2O(l) ---> H2SO4(aq)
H2SO4(concetrated)+Ba(OH)2(aq) ---> BaSO4(s) + 2H2O(s)
After fully reacted oleum with Ba(OH)2 excess we received BaSO4 precipitate which were filtrated, dryed and weighed. We had 2,54 grams dry BaSO4 salt. Calculate coefficient x value in oleum formula H2SO4*xSO3. Write a consistent answer.
1
Expert's answer
2019-10-17T07:19:09-0400

Moles of "BaSO_4"


"n=\\frac{m}{M} = \\frac{2.54}{233.39} =0.0109 mol"

As 1 mole of "BaSO_4" contains 1 mole of S , then n(S) = 0.0109 mol

Let the formula of oleum to be "H_2SO_4\\times aSO_3"

Let moles of "H2SO_4" in oleum to be x moles, and of SO3 to be y moles


Then 1 gram of oleum is


"m(H_2SO_4) + m(SO_3)=1"


"98.08x + ay80.08 = 1"

Also we know the total moles of S


"x+ay =0.0109"

Solve the system of these two equations:

If a=1 then y= 0.0038, x = 0.0071

If a=2 then y=0.0019, x = 0.0071

If a=3 then y = 0.0013, x = 0.0071

The value of a can be different (1,2,3.....). All these answers are appropriate. It seems like there is lack of data to give accurate answer.


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