# Question 2270c

Aug 10, 2017

$8.43 \times {10}^{24}$ $\text{atoms C}$

#### Explanation:

We're asked to find the number of atoms of $\text{C}$ necessary to produce $12.0$ $\text{mol CO}$, given the reaction is

$2 \text{B"_2"O"_3(s) + 7"C"(s) rarr "B"_4"C"(s) + 6"CO} \left(g\right)$

First, let's use the coefficients of the chemical equation to find the relative number of moles of $\text{C}$ that react:

12.0cancel("mol CO")((7color(white)(l)"mol C")/(6cancel("mol CO"))) = color(red)(ul(14.0color(white)(l)"mol C"#

Now, we'll use Avogadro's number ($6.022 \times {10}^{23}$ ${\text{mol}}^{-} 1$) to convert from moles to atoms:

$\textcolor{red}{14.0} \cancel{\textcolor{red}{\text{mol C"))((6.022xx10^23color(white)(l)"atoms C")/(1cancel("mol C"))) = color(blue)(ulbar(|stackrel(" ")(" "8.43xx10^24color(white)(l)"atoms C"" }} |}$

rounded to $3$ significant figures, the amount allowed in the problem.