Question

Question #ffa43

Answer 1

`"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 `color(blue)("328.24 g")` `"mol"^(-1)`, which means that every `1` mole of cerium(III) oxide has a mass of `color(blue)("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 color(red)(cancel(color(black)("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.