Question

What is the volume of 75.0 g of `O_2` at STP?

Answer 1

`V=105L`

Explanation:

`p=1 atm`
`t=273.15 K`
`R=0.08206 (L*atm)/(mol*K)`
`n=(75.0gO_2)/(16.00(g/(mol)O_2)=4.6875mol`

Ideal gas law: `PV=nRT`, solve for volume:
`V=(nRT)/P`

`V=(4.6875mol*0.08206 (L*atm)/(mol*K)*273.15K)/(1atm)`

`V=105 L`