To determine the number of molecules of [tex]\(CO_2\)[/tex] produced when 2.00 grams of [tex]\(C_2H_5OH\)[/tex] (ethanol) are produced, we need to follow these steps:
1. Determine the molar mass of [tex]\(C_2H_5OH\)[/tex]:
- The molar mass of [tex]\(C\)[/tex] (carbon) is approximately 12.01 g/mol.
- The molar mass of [tex]\(H\)[/tex] (hydrogen) is approximately 1.01 g/mol.
- The molar mass of [tex]\(O\)[/tex] (oxygen) is approximately 16.00 g/mol.
- Therefore, the molar mass of [tex]\(C_2H_5OH\)[/tex] is calculated as:
[tex]\[
(2 \times 12.01) + (6 \times 1.01) + (1 \times 16.00) = 24.02 + 6.06 + 16.00 = 46.08 \, \text{g/mol}
\][/tex]
For the purpose of this calculation, we'll use a rounded value of 46.07 g/mol.
2. Calculate the number of moles of [tex]\(C_2H_5OH\)[/tex] produced:
- Using the formula:
[tex]\[
\text{Number of moles} = \frac{\text{mass}}{\text{molar mass}}
\][/tex]
- Inserting the values:
[tex]\[
\text{Number of moles of } C_2H_5OH = \frac{2.00 \, \text{g}}{46.07 \, \text{g/mol}} \approx 0.04341 \, \text{moles}
\][/tex]
3. Determine the moles of [tex]\(CO_2\)[/tex] produced:
- According to the balanced chemical equation, 1 mole of [tex]\(C_2H_5OH\)[/tex] produces 1 mole of [tex]\(CO_2\)[/tex].
- Therefore, the number of moles of [tex]\(CO_2\)[/tex] produced is the same as the number of moles of [tex]\(C_2H_5OH\)[/tex] produced.
[tex]\[
\text{Number of moles of } CO_2 = 0.04341 \, \text{moles}
\][/tex]
4. Calculate the number of molecules of [tex]\(CO_2\)[/tex] produced:
- Using Avogadro’s number, which is approximately [tex]\(6.022 \times 10^{23}\)[/tex] molecules/mol.
- The formula to calculate the number of molecules is:
[tex]\[
\text{Number of molecules} = \text{Number of moles} \times \text{Avogadro's number}
\][/tex]
Substituting the values:
[tex]\[
\text{Number of molecules of } CO_2 = 0.04341 \, \text{moles} \times 6.022 \times 10^{23} \, \text{molecules/mol} \approx 2.614 \times 10^{22} \, \text{molecules}
\][/tex]
Therefore, the number of molecules of [tex]\(CO_2\)[/tex] formed when 2.00 g of [tex]\(C_2H_5OH\)[/tex] are produced is approximately [tex]\(2.614 \times 10^{22}\)[/tex] molecules.
