Question #9f4fc

1 Answer
Aug 10, 2017

Answer:

#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"" ")|)#