Question

For the following reaction, 23.7 grams of Sulfur Dioxide are alowed to react with 9.57 grams

of Oxygen

Sulfur Dioxide+Oxygen====>Sulfur Trioxide

Part A. What is the maximum amount of Sulfur Trioxide that can be formed?

Part B. What is the forumla for the Limiting Reagent?

Part C. What amount of excess reagent remains after the reaction is

complete?

Answer 1

equation of the reaction: 2 SO2 + O2 2 SO3

Mole sof SO2 = mass / molar mass = 23.7 / 64 = 0.37 moles

Moles of O2 = mass / molar mass = 9.57 / 32 = 0.30 moles

Since moles of O2 needed = moles of SO2, we have fewer moles of O2 meaning it is the limiting reagent.

Amount of SO2 reacted = moles of O2 = 0.37 - 0.30 = 0.07 moles

mass of SO2 left = moles * molar mass = 0.07 * 64 = 4.48 g