For the following reaction, 23.9 grams of sulfur dioxide are

Last updated: 8/3/2022

For the following reaction, 23.9 grams of sulfur dioxide are allowed to react with 9.27 grams of oxygen gas. sulfur dioxide (g) + oxygen (g) ---> sulfur trioxide (g) What is the maximum amount of sulfur trioxide that can be formed? What is the FORMULA for the limiting reagent? What amount of the excess reagent remains after the reaction is complete? 46.35