Question

How many grams of oxygen gas are needed to react completely with 38.442 grams of sulfur dioxide gas?

Answer 1

Answer: 19.221 grams of oxygen gas are needed to react completely with 38.442 grams of sulfur dioxide gas.

Explanation:

Given :$SO_2(g)+O_2(g)\rightarrow SO_3(g)$

After balancing, $2SO_2(g)+O_2(g)\rightarrow 2SO_3(g)$

$\text{Number of moles}SO_2=\frac{\text{mass of the compound}}{\text{Molecular mass of the compound}}=\frac{38.442 g}{32g/mol}=1.2013 moles$

According to reaction ,2 moles of $SO_2$ reacts with one mole of $O_2$ then 1.2013 moles of $SO_2$ will react with :

$=\frac{1}{2}\times 1.2013\text{moles of}O_2=0.6006 moles$

Mass of $O_2$ required =$\text{Number of moles of compound}\times \text{Molecular mass of the compound}=0.6006 moles\times 32g/mol=19.221 g$

19.221 grams of oxygen gas are needed to react completely with 38.442 grams of sulfur dioxide gas.