# According to the following reaction, how many moles of sulfur trioxide will be formed upon the complete reaction of 0.971 moles of sulfur dioxide with excess oxygen gas? sulfer dioxide (g) + oxygen (g) -> sulfer trioxide

Nov 6, 2016

Surely it will be $0.971 \cdot m o l$ of $S {O}_{3}$?
$S {O}_{2} \left(g\right) + \frac{1}{2} {O}_{2} \left(g\right) \rightarrow S {O}_{3} \left(g\right)$
The equation unequivocally tells us that the reaction of $64 \cdot g$ $S {O}_{2} \left(g\right)$ with $16 \cdot g$ ${O}_{2} \left(g\right)$ gives $80 \cdot g$ $S {O}_{3} \left(g\right)$. The given masses are the molar equivalents of each gas. Again, from the stoichometric equation, sulfur was the limiting reagent (because excess dioxygen gas was specified); there was a given molar quantity of sulfur (as its dioxide), and thus a given molar quantity of the trioxide.