Question

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.

Answer 1

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.