Let's break down the solution step-by-step with all the calculations.
1. Convert the mass of carbon monoxide (CO) from milligrams to grams:
Given mass: 280 mg
Conversion:
[tex]\[
280 \, \text{mg} = \frac{280}{1000} \, \text{g} = 0.28 \, \text{g}
\][/tex]
2. Determine the molar mass of CO:
- Carbon (C) has a molar mass of 12 g/mol.
- Oxygen (O) has a molar mass of 16 g/mol.
Therefore, the molar mass of CO:
[tex]\[
\text{Molar mass of CO} = 12 \, \text{g/mol} + 16 \, \text{g/mol} = 28 \, \text{g/mol}
\][/tex]
3. Calculate the number of moles of CO in 280 mg (0.28 g):
Use the formula:
[tex]\[
\text{Number of moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}}
\][/tex]
[tex]\[
\text{Number of moles of CO} = \frac{0.28 \, \text{g}}{28 \, \text{g/mol}} = 0.01 \, \text{moles}
\][/tex]
4. Calculate the number of moles of CO in [tex]$2 \times 10^{21}$[/tex] molecules:
Given:
- Avogadro's number = [tex]\(6.022 \times 10^{23}\)[/tex] molecules/mole
Use the formula to find the number of moles from molecules:
[tex]\[
\text{Number of moles} = \frac{\text{number of molecules}}{\text{Avogadro's number}}
\][/tex]
[tex]\[
\text{Number of moles removed} = \frac{2 \times 10^{21} \, \text{molecules}}{6.022 \times 10^{23} \, \text{molecules/mol}} \approx 0.003321 \, \text{moles}
\][/tex]
5. Determine the number of moles of CO remaining after removal:
Subtract the number of moles removed from the initial number of moles:
[tex]\[
\text{Number of moles left} = 0.01 \, \text{moles (initial)} - 0.003321 \, \text{moles (removed)} = 0.006679 \, \text{moles}
\][/tex]
Therefore, after removing [tex]$2 \times 10^{21}$[/tex] molecules from 280 mg of CO, the number of moles of CO left is approximately [tex]\(0.006679\)[/tex] moles.
