# How many moles are in 98.3 grams of aluminum hydroxide, Al(OH)_3?

$\text{Number of moles}$ $=$ $\text{Mass"/"Molar mass}$
$\text{Number of moles, } A l {\left(O H\right)}_{3}$ $=$ $\frac{98.3 \cdot g}{78.0 \cdot g \cdot m o {l}^{-} 1}$
$=$ $\text{A bit over 1 mole...........}$