# How many moles of air are in a 6.06-L tire at STP?

The molar volume at STP is $22.4 \cdot L$
And thus at $\text{STP}$, $6.06 \cdot L$ represents $\frac{6.06 \cdot L}{22.4 \cdot L \cdot m o {l}^{-} 1}$ $=$ ??*mol, $\cong \frac{1}{4} \cdot m o l$.