# 1 xx 10^21 molecules are removed from 280" mg" of carbon monoxide. Calculate the number of moles of carbon monoxide left? [Answer:8.4 xx 10^- 3" moles"]

Jun 13, 2018

Approximately $8.4$ millimoles

#### Explanation:

Total moles of Carbon monoxide ($\text{CO}$) is

$\text{n"_"Total" = (280 × 10^-3\ "g")/"28 g/mol" = 10^-2\ "mol}$

Moles of $\text{CO}$ removed is

"n"_"removed" = 1 × 10^21 cancel"molecules" × "1 mol"/(6.022 × 10^23 cancel"molecules")

color(white)("n"_"removed") = 1.66 × 10^-3\ "mol"

Moles of $\text{CO}$ left is

$\text{n"_"left" = "n"_"Total" - "n"_"removed}$

$\textcolor{w h i t e}{\text{n"_"left") = 10^-2\ "mol" - (1.66 × 10^-3\ "mol}}$

color(white)("n"_"left") = (10 - 1.66) × 10^-3\ "mol"

color(white)("n"_"left") ≈ 8.4 × 10^-3\ "mol"

Jun 13, 2018

$8 , 34 \cdot {10}^{-} 3$ moles

#### Explanation:

CO has a molecular mass of 28, so 280 mg are 10 millimoles.
${10}^{21}$ molecules are${10}^{21} / {N}_{A} = 1.66 \cdot {10}^{-} 3$ moles,
Subtracting $1.66 \cdot {10}^{-} 3$ moles from$10 \cdot {10}^{-} 3$ moles gives $8.34 \cdot {10}^{-} 3$ moles