Answer:
[tex]10.6g[/tex] of [tex]SO_{2}[/tex]
[tex]2.0g[/tex] of [tex]H_{2}O[/tex]
Explanation:
First we are going to balance the reaction:
[tex]S+_{2}H_{2}SO{4}=_{3}SO_{2}+_{2}H_{2}O[/tex]
Then we are going to find the quantities with the stoichiometry:
-For [tex]SO_{2}[/tex]:
[tex]10.8gH_{2}SO_{4}*\frac{1molH_{2}SO_{4}}{98.079gH_{2}SO_{4}}*\frac{3molesSO_{2}}{2molesH_{2}SO_{4}}*\frac{64.066gSO_{2}}{1molSO_{2}}=10.6gSO_{2}[/tex]
-For [tex]H_{2}O[/tex]:
[tex]10.8gH_{2}SO_{4}*\frac{1molH_{2}SO_{4}}{98.079gH_{2}SO_{4}}*\frac{2molesH_{2}O}{2molesH_{2}H_{2}O}*\frac{18.015gH_{2}O}{1molH_{2}O}=10.6gH_{2}O=2.0gH_{2}O[/tex]