Question

What volume does 22.4 g of chlorine gas occupy at STP?

Answer 1

`7.08 L`

Explanation:

Because we are at STP, we must use the ideal gas law equation
`PxxV = nxxRxxT`.

• P represents pressure (must have units of atm)
• V represents volume (must have units of liters)
• n represents the number of moles
• R is the proportionality constant (has units of `(Lxxatm)/ (molxxK)`)
• T represents the temperature, which must be in Kelvins.

Next, list your known and unknown variables. Our only unknown is the volume of `Cl_2(g)`. Our known variables are P,n,R, and T.

At STP, the temperature is 273K and the pressure is 1 atm. The proportionality constant, R, is equal to 0.0821 `(Lxxatm)/ (molxxK)`
The only issue is the mass of `Cl_2(g)`, we need to convert it into moles of `Cl_2(g)` in order to use the formula.
`22.4cancel"g" xx (1molCl_2)/(70.90cancel"g")` = 0.316 mol `Cl_2`

Now all we have to do is rearrange the equation and solve for V like so:
`V = (nxxRxxT)/P`
`V = (0.316molxx0.0821(Lxxatm)/(molxxK)xx(273K))/(1atm)`
`V = 7.08 L`