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?

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.

the answer to your question is c.

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