Question #5db3b

1 Answer
anor277
May 2, 2016

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#

Related questions
See all questions in Equation Stoichiometry
Impact of this question
2286 views around the world
You can reuse this answer
Creative Commons License