The reaction of sulfur with oxygen is shown below. 2s(s) + 3o2(g) → 2so3(g) calculate the mass of sulfur trioxide (in g) produced when 100.0 g of each reactant is present.

The mass of sulfur trioxide produced will be 167 grams

Explanation:

To calculate the number of moles, we use the equation:

     .....(1)

Given mass of sulfur = 100.0 g

Molar mass of sulfur = 32.07 g/mol

Putting values in equation 1, we get:

Given mass of oxygen gas = 100.0 g

Molar mass of oxygen gas = 32 g/mol

Putting values in equation 1, we get:

The given chemical equation follows:

By Stoichiometry of the reaction:

3 moles of oxygen gas reacts with 2 moles of sulfur metal

So, 3.125 moles of oxygen gas will react with = of sulfur metal

As, given amount of sulfur metal is more than the required amount. So, it is considered as an excess reagent.

Thus, oxygen gas is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

3 moles of oxygen gas produces 2 moles of sulfur trioxide.

So, 3.125 moles of oxygen gas will produce of sulfur trioxide

Now, calculating the mass of sulfur trioxide from equation 1, we get:

Molar mass of sulfur trioxide = 80.1 g/mol

Moles of sulfur trioxide = 2.08 moles

Putting values in equation 1, we get:

Hence, the mass of sulfur trioxide produced will be 167 grams