Question #9f4fc

1 Answer
Aug 10, 2017

5.965.96 "g B"_4"C"g B4C

Explanation:

We're asked to find the mass of "B"_4"C"B4C formed if 14.514.5 "L CO"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 22.422.4 "L"L.

Therefore, let's use this statement to covert from liters of "CO"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"" ")|)