Question

After going through a guided tutorial by selecting Run Grams Demonstration, you can create your own experiment by clicking the Run Experiment button at the end or by clicking the Overview tab and returning to the Experiment tab to select Run Experiment. There are nine reactions you can explore on your own. Sulfur dioxide gas (SO2) and oxygen gas (O2) react to form the liquid product of sulfur trioxide (SO3). How much SO2 would you need to completely react with 6.00 g of O2 such that all reactants could be consumed

Answer 1

Answer: Thus 24.0 g of
`SO_2` would be needed.

Step-by-step explanation:

To calculate the moles :

`\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}`

`\text{Moles of} O_2=(6.00g)/(32g/mol)=0.1875moles`

`2SO_2(g)+O_2(g)\rightarrow 2SO_3(l)`

According to stoichiometry :

1 mole of
`O_2` require = 2 moles of
`SO_2`

Thus 0.1875 moles of
`O_2` will require=
`(2)/(1)* 0.1875=0.375moles` of
`SO_2`

Mass of
`SO_2=moles* {\text {Molar mass}}=0.375moles* 64g/mol=24.0g`

Thus 24.0 g of
`SO_2` would be needed to completely react with 6.00 g of
`O_2` such that all reactants could be consumed.