Get the answers you need at Westonci.ca, where our expert community is dedicated to providing you with accurate information. Experience the convenience of getting reliable answers to your questions from a vast network of knowledgeable experts. Join our platform to connect with experts ready to provide precise answers to your questions in different areas.
Sagot :
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].
[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].
Thank you for trusting us with your questions. We're here to help you find accurate answers quickly and efficiently. We hope this was helpful. Please come back whenever you need more information or answers to your queries. Your questions are important to us at Westonci.ca. Visit again for expert answers and reliable information.