Question

Question #9f4fc

Answer 1

`5.96` `"g B"_4"C"`

Explanation:

We're asked to find the mass of `"B"_4"C"` formed if `14.5` `"L CO"` are produced at S.T.P.

To do this, we'll recognize a fact that most chemists agree by (although it depends on the values chosen for standard temperature and pressure):

One mole of an (ideal) gas at S.T.P. occupies a volume of `22.4` `"L"`.

Therefore, let's use this statement to covert from liters of `"CO"` to moles**:

`14.5cancel("L CO")((1color(white)(l)"mol CO")/(22.4cancel("L CO"))) = color(red)(ul(0.647color(white)(l)"mol CO"`

Now, we'll use the coefficients of the chemical equation to find the relative number of moles of `"B"_4"C"` that react:

`color(red)(0.647)cancel(color(red)("mol CO"))((1color(white)(l)"mol B"_4"C")/(6cancel("mol CO"))) = color(green)(ul(0.108color(white)(l)"mol B"_4"C"`

And finally, we'll use the molar mass of `"B"_4"C"` (`55.255` `"g/mol"`) to find the number of grams that react:

`color(green)(0.108)cancel(color(green)("mol B"_4"C"))((55.255color(white)(l)"g B"_4"C")/(1cancel("mol B"_4"C"))) = color(blue)(ulbar(|stackrel(" ")(" "5.96color(white)(l)"g B"_4"C"" ")|)`