Answer
8.15 g
Work Step by Step
Let's take,
Initial number of moles = $n_{i}$
Final number of moles = $n_{f}$
The mass of removed = (Number of moles withdrawn)(Mass per mole)
Nitrogen
$m=(n_{i}-n_{f})(Mass\space per\space mole)-(1)$
Let's apply the ideal gas law $PV=nRT$ to find the number of initial and final moles.
$PV=nRT=>n=\frac{PV}{RT}$
$n_{f}=\frac{P_{f}V}{RT}$ and $n_{i}=\frac{P_{i}V}{RT}$
Therefore, we can get,
$\frac{n_{f}}{n_{i}}=\frac{\frac{P_{f}V}{RT}}{\frac{P_{i}V}{RT}}=\frac{P_{f}}{P_{i}}=>n_{f}=n_{i}(\frac{P_{f}}{P_{i}})-(2)$
(2)=>(1),
$m=\{n_{i}-n_{i}(\frac{P_{f}}{P_{i}})\}(Mass\space per\space mole)$
Let's plug known values into this equation.
$m=\{0.85\space mol-(0.85\space mol)(\frac{25\space atm}{38\space atm})\}(28.0134\space g/mol)=8.15\space g$
Mass of the removed nitrogen = 8.15 g
