# How do you convert 9.45 moles g of Al_2O_3 into grams of Al_2O_3?

Dec 9, 2015

$963.9 g$

#### Explanation:

consider the relationship between the number of moles, molar mass and the mass of a compound,

$n = \frac{m}{M}$

where n is the number of moles, m is the mass of the compound and M is the molar mass of the compound.

you just have to multiply the number of moles (9.45 moles) by the molar mass of $A {l}_{\text{2"O_"3}}$ which is $27 \cdot 2 + 16 \cdot 3 = 102$

so,
$m = 9.45 m o l \cdot 102 g m o {l}^{\text{-1}}$

$m = 963.9 g$