According to the following reaction, how many grams of sulfur trioxide will be formed upon the complete reaction of 21.2 grams of oxygen gas with excess sulfur dioxide?

sulfur dioxide (g) + oxygen (g) sulfur trioxide (g)

Here is an example stoichiometry problem I've posted. Just follow the steps.

http://www.jiskha.com/science/chemistry/stoichiometry.html

According to the following reaction, how many grams of water will be formed upon the complete reaction of 21.6 grams of oxygen gas with excess hydrogen sulfide?

2 H2S (g) + 3 O2 (g) = 2 H2O (l) + 2 SO2 (g)

To determine the amount of sulfur trioxide formed in the reaction, you need to use the concept of stoichiometry.

First, balance the equation to ensure that the number of atoms on both sides of the reaction is equal:

2 SO2(g) + O2(g) 2 SO3(g)

From the balanced equation, you can see that 2 moles of sulfur dioxide react with 1 mole of oxygen gas to produce 2 moles of sulfur trioxide.

To solve the problem, you can follow the following steps:

Step 1: Convert the given mass of oxygen gas to moles.
- To do this, you need to know the molar mass of oxygen, which is approximately 32 grams per mole.
- Divide the given mass of oxygen gas (21.2 grams) by the molar mass of oxygen to obtain the number of moles.

Step 2: Use the balanced equation to determine the stoichiometric ratio between oxygen and sulfur trioxide.
- The ratio from the balanced equation is 1 mole O2: 2 moles SO3.
- Since you have the number of moles of oxygen, you can determine the number of moles of sulfur trioxide.

Step 3: Convert moles of sulfur trioxide to grams.
- Determine the molar mass of sulfur trioxide, which is approximately 80 grams per mole.
- Multiply the number of moles of sulfur trioxide by its molar mass to get the mass in grams.

By following these steps, you can calculate the amount of sulfur trioxide formed upon the complete reaction of 21.2 grams of oxygen gas with excess sulfur dioxide.