# Question #ffa43

Jul 8, 2017

$\text{131 g}$

#### Explanation:

The idea here is that you can convert between the number of grams present in a sample of cerium(III) oxide and the number of moles it contains by using the compound's molar mass.

Cerium(III) oxide has a molar mass of $\textcolor{b l u e}{\text{328.24 g}}$ ${\text{mol}}^{- 1}$, which means that every $1$ mole of cerium(III) oxide has a mass of $\textcolor{b l u e}{\text{328.24 g}}$.

You can thus use the molar mass of the compound as a conversion factor to determine the number of grams of cerium(III) oxide that would contain $0.400$ moles.

$0.400 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles Ce"_2"O"_3))) * color(blue)("328.24 g")/(1color(red)(cancel(color(black)("mole Ce"_2"O"_3)))) = color(darkgreen)(ul(color(black)("131 g}}}}$

The answer is rounded to three sig figs, the number of significant figures you have for the number of moles present in the sample.