Question

Question #5db3b

Answer 1

Approx. `15*g`

Explanation:

`SO_2(g) + 1/2O_2 rarr SO_3(g)`

`"Moles of sulfur dioxide"` `=` `(12.4*g)/(64.06*g*mol^-1)` `=` `0.194*mol`.

`"Moles of dioxygen"` `=` `(3.45*g)/(32.00*g*mol^-1)` `=` `0.108*mol`.

Clearly, from the molar quantities there is a slight excess of oxygen. And since, to start, there are `0.194*mol` `SO_2`, to finish there would be an equimolar quantity of `SO_3(g)` given quantitative yield.

`"Mass of " SO_3` `=` `0.194*molxx80.07*g*mol^-1=15.5*g`