# What is the mass in grams contained in 99.02 L of NO2 at STP?

Dec 12, 2014

The answer is $203 g$.

At STP - Standard Temperature and Pressure (273.15K, 1 atm) - an ideal gas' molar volume is equal to $22.4 L$. This means that $1$ mole of any ideal gas always occupies that volume at that pressure and that temperature.

Now, even before calculating the actual value, one must try and estimate what that value could be; since we were given $99.02 L$, it's fairly easy to see that we must be dealing with more than $1$ mole.

SInce 22.4 is between 4 and 5 times smaller (more than 4, less than 5) than 99.02, we could estimate the number of moles to be 4.5, which, given $N {O}_{2}$'s molar mass of $46 \frac{g}{m o l}$, would result in 207g.

The actual number of moles is

${n}_{N {O}_{2}} = \frac{V}{{V}_{m o l a r}} = \frac{99.02 L}{22.4 L} = 4.42$ moles, which means

${m}_{N {O}_{2}} = n \cdot m o l a r m a s s = 4.42 m o l e s \cdot 46 \frac{g}{m o l e} = 203 g$.

Notice the similarity between the estimate and the actual result...