Question

32 g of sulfur will react with 48 g of oxygen to produce 80 g of sulfur trioxide. if 20 g of sulfur and 100 g of oxygen are placed into a sealed container and allowed to react, what is the mass of sulfur trioxide in the container after the reaction is completed?

Answer 1

The balanced chemical equation for the reaction of sulfur and oxygen to form sulfur trioxide is as follows:

2S+3O_{2}\rightarrow 2SO_{3}

That means 2 moles of sulfur reacts with 3 moles of oxygen to give 2 moles of sulfur trioxide.

The reaction is always measured in terms of number of moles.

First check for the limiting reagent here, the mole ratio is 1 S:1 SO_{3} and 1.5O_{2}:1SO_{3}

Here, oxygen is in excess thus, sulfur is limiting reactant.

Now, here, 20 g of sulfur and 100 g of oxygen is placed in the container. First convert the mass into moles as follows:

For sulfur: molar mass is 32 g/mol

Thus,

n=\frac{m}{M}=\frac{20 g}{32 g}=0.625 mol

Now, 1 mol of S gives 1 mol of SO_{3} thus, 0.625 mol of S gives 0.625 mol of SO_{3}.

calculate mass of SO_{3} produced from molar mass and number of moles as follows:

m=n×M=0.625 mol×80 g/mol=50 g

Thus, mass of sulfur trioxide produced will be 50 g.