# How many moles of sodium bicarbonate, NaHCO_3, are in 508g, of NaHCO_3?

Almost $6 \cdot m o l$.
The molar mass of $N a H C {O}_{3}$ $=$ $84.01 \cdot g \cdot m o {l}^{-} 1$.
Thus, $\text{moles of sodium bicarbonate}$ $=$ $\text{Mass"/"Molar Mass}$
$=$ $\frac{508 \cdot \cancel{g}}{84.01 \cdot \cancel{g} \cdot m o {l}^{-} 1}$ $=$ ??mol