Hi, kimchi2755 :)

As you mentioned, mass must be conserved, yes, that's true. 

However, the mass of each reactant is not necessarily mixed as much as they react. That's why we learn about the limiting reactant and the excess reactant. 

This time, 2SO2 + O2 --> 2SO3, which means 2 moles of SO2 react with 1 mol of O2, right? 

90g of SO2 represents 1.4 mol (90/64) and 100 g of O2 represents 3.125 mol. 

(you said 100 g but i think you meant this questions about the reaction between 90 g of SO2 and 100 g of O2, right?) 

So, it's already not representing a proper ratio of 2 to 1 between SO2 and O2. 

Thereby, SO2 is the limiting reactant. When 1.4 mol of SO2 is used up, that also uses 0.7 mol of O2 up. Thus, 1.4 mol of SO3 will be produced. 

1.4 mol SO3 * 80 g/mol = 112 g of SO3, right? 

But there's the excess amount of O2 (3.125-0.7)mol = 2.425 mol *32g/mol = 77.6 g 

So, it's like this, none of SO2 remains but 77.6 g of O2 remains as unreacted and there's a product of SO3 by 112 g 

In the beginning, 90 g SO2 + 100 g O2 = 190 g 

And 78 g O2 + 112 g SO3 = 190 g 

Mass is conserved anyways, right? 

I hope you understand thorougly now. 

:) 

Kelda