Certainly! Let's consider the reaction:
[tex]\[ 2 \text{Mg}(s) + \text{O}_2(g) \rightarrow 2 \text{MgO}(s) \][/tex]
We need to find the volume of oxygen gas ([tex]\(\text{O}_2\)[/tex]) required to react with [tex]\(4.03 \, \text{g}\)[/tex] of magnesium ([tex]\(\text{Mg}\)[/tex]) at standard temperature and pressure (STP).
### Step-by-Step Solution:
1. Determine the moles of magnesium ([tex]\(\text{Mg}\)[/tex]):
- Given mass of [tex]\(\text{Mg}\)[/tex]: [tex]\(4.03 \, \text{g}\)[/tex]
- Molar mass of [tex]\(\text{Mg}\)[/tex]: [tex]\(24.305 \, \text{g/mol}\)[/tex]
- Moles of [tex]\(\text{Mg}\)[/tex]:
[tex]\[
\text{moles of Mg} = \frac{\text{mass of Mg}}{\text{molar mass of Mg}} = \frac{4.03 \, \text{g}}{24.305 \, \text{g/mol}} \approx 0.1658 \, \text{mol}
\][/tex]
2. Determine the moles of oxygen gas ([tex]\(\text{O}_2\)[/tex]) needed:
- According to the reaction stoichiometry, [tex]\(2 \, \text{moles of Mg}\)[/tex] react with [tex]\(1 \, \text{mole of O}_2\)[/tex].
- Therefore, moles of [tex]\(\text{O}_2\)[/tex] required:
[tex]\[
\text{moles of O}_2 = \frac{\text{moles of Mg}}{2} = \frac{0.1658 \, \text{mol}}{2} \approx 0.0829 \, \text{mol}
\][/tex]
3. Convert moles of oxygen gas to volume at STP:
- At STP, [tex]\(1 \, \text{mole of gas}\)[/tex] occupies [tex]\(22.414 \, \text{L}\)[/tex].
- Volume of [tex]\(\text{O}_2\)[/tex] in liters:
[tex]\[
\text{volume of O}_2 \, (\text{L}) = \text{moles of O}_2 \times 22.414 \, \text{L/mol} \approx 0.0829 \, \text{mol} \times 22.414 \, \text{L/mol} \approx 1.858 \, \text{L}
\][/tex]
4. Convert the volume from liters to milliliters:
- [tex]\(1 \, \text{L} = 1000 \, \text{mL}\)[/tex]
- Volume of [tex]\(\text{O}_2\)[/tex] in milliliters:
[tex]\[
\text{volume of O}_2 \, (\text{mL}) = 1.858 \, \text{L} \times 1000 \, \text{mL/L} \approx 1858 \, \text{mL}
\][/tex]
Thus, the volume of oxygen gas required to react with [tex]\(4.03 \, \text{g}\)[/tex] of magnesium at STP is approximately [tex]\(1858 \, \text{mL}\)[/tex].
In the given choices, the closest value is:
[tex]\(1860 \, \text{mL}\)[/tex].
So, the correct answer is [tex]\(1860 \, \text{mL}\)[/tex].
