Question

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

SO2(g) + O2(g) --> SO3(g) (needs to be balanced)

Answer 1

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

Step-by-step 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}}=(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 :

`=(1)/(2)* 1.2013\text{moles of}O_2=0.6006 moles`

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

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