Question

What is the the volume occupied by 2.34 g of carbon dioxide gas at STP?

Answer 1

`V = 1.18 L`

Explanation:

In order to solve this problem we would use the Ideal Gas Law formula `PV=nRT`

`P =` Pressure in `atm`
`V=` Volume in `L`
`n=` moles
`R=` Ideal Gas Law Constant
`T=` Temp in `K`

`STP` is Standard Temperature and Pressure which has values of
`1 atm` and `273K`

`2.34g CO_2` must be converted to moles

`2.34g CO_2 x (1mol)/(44gCO_2) = 0.053 mols`

`P = 1atm`
`V= ??? L`
`n= 0.053 mols`
`R=0.0821 (atmL)/(molK)`
`T=273K`

`PV=nRT` becomes

`V = (nRT)/P`

`V = (0.053cancel(mols)(0.0821 (cancel(atm)L)/(cancel(mol)cancel(K)))(273cancelK))/(1atm)`

`V = 1.18 L`

Answer 2

`1.18 L`

Explanation:

Alternative solution:

Calculate how many moles `2.34g` `CO_2` is

`2.34g * (1mol)/(44g CO_2) = 0.053 mol`

Simple mole ratios tell us that `1` mol of any gas at STP is `22.4L`, but since we only have `0.053 mol`, we times `22.4L/(mol)` by `0.053mol.`

`(22.4L)/cancel(mol) * 0.053 cancel(mol) = 1.18L`