# Question 9f4fc

Aug 10, 2017

$5.96$ $\text{g B"_4"C}$

#### Explanation:

We're asked to find the mass of $\text{B"_4"C}$ formed if $14.5$ $\text{L CO}$ 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 $22.4$ $\text{L}$.

Therefore, let's use this statement to covert from liters of $\text{CO}$ 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 $\text{B"_4"C}$ 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 $\text{B"_4"C}$ ($55.255$ $\text{g/mol}$) to find the number of grams that react:

$\textcolor{g r e e n}{0.108} \cancel{\textcolor{g r e e n}{\text{mol B"_4"C"))((55.255color(white)(l)"g B"_4"C")/(1cancel("mol B"_4"C"))) = color(blue)(ulbar(|stackrel(" ")(" "5.96color(white)(l)"g B"_4"C"" }} |}$