# How do you convert 64.3 grams of Al_2(CO_3)_3 to moles?

$\text{Moles}$ $=$ $\text{Mass of stuff"/"Molar mass of stuff}$
So $\frac{64.3 \cdot \cancel{g}}{233.99 \cdot \cancel{g} \cdot m o {l}^{-} 1}$ $\cong$ $0.25$ $m o l$.