# How many moles of aluminum oxide are in 6.83 g Al_2O_3?

Approx. $0.07$ moles.
Moles of $A {l}_{2} {O}_{3}$ $=$ $\frac{6.83 \cdot \cancel{g}}{101.96 \cdot \cancel{g} \cdot m o {l}^{-} 1}$ $=$ ?? $m o l$ $A {l}_{2} {O}_{3}$
Please note that the answer is consistent dimensionally. Setting up the problem like this gives an answer in ${\left(m o {l}^{-} 1\right)}^{-} 1$ $=$ $\text{moles}$ AS REQUIRED.