Answered

Consider this reaction:

[tex]\[ 2 \, \text{Mg} (s) + \, \text{O}_2 (g) \rightarrow 2 \, \text{MgO} (s) \][/tex]

What volume (in milliliters) of oxygen gas is required to react with \(4.03 \, \text{g}\) of \( \text{Mg} \) at STP?

A. \(1850 \, \text{mL}\)

B. \(2880 \, \text{mL}\)

C. \(3710 \, \text{mL}\)

D. [tex]\(45,100 \, \text{mL}\)[/tex]



Answer :

GinnyAnswer
To determine the volume of oxygen gas required to react with \(4.03\) grams of magnesium (\(\text{Mg}\)) at standard temperature and pressure (STP), we need to follow a step-by-step process involving stoichiometric calculations and gas laws. Here’s the detailed solution:

1. Determine moles of magnesium:
- Molar mass of magnesium (\(\text{Mg}\)) \(\approx 24.305 \ \text{g/mol}\)
- Given mass of magnesium (\(\text{Mg}\)) = \(4.03 \ \text{g}\)
- Moles of \(\text{Mg}\) can be calculated using the formula:
[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. Use the balanced chemical equation to find moles of \(\text{O}_2\):
- The balanced equation is:
[tex]\[ 2 \text{Mg} (s) + \text{O}_2 (g) \rightarrow 2 \text{MgO} (s) \][/tex]
- According to the balanced equation, \(2\) moles of \(\text{Mg}\) react with \(1\) mole of \(\text{O}_2\).
- Therefore, moles of \(\text{O}_2\) needed can be calculated as:
[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. Find the volume of \(\text{O}_2\) gas at STP:
- At STP, \(1 \ \text{mol}\) of any gas occupies a volume of \(22.414 \ \text{L}\) (standard molar volume).
- Volume of \(\text{O}_2\) in liters can be calculated using the formula:
[tex]\[ \text{volume of O}_2 (\text{L}) = \text{moles of O}_2 \times 22.414 \ \text{L/mol} = 0.0829 \ \text{mol} \times 22.414 \ \text{L/mol} \approx 1.858 \ \text{L} \][/tex]

4. Convert the volume from liters to milliliters:
- Since \(1 \ \text{L} = 1000 \ \text{mL}\), the volume in milliliters is:
[tex]\[ \text{volume of O}_2 (\text{mL}) = 1.858 \ \text{L} \times 1000 \ \text{mL/L} \approx 1858.2 \ \text{mL} \][/tex]

Hence, the volume of oxygen gas required to react with \(4.03 \ \text{g}\) of magnesium at STP is approximately \(1858.2 \ \text{mL}\). The closest match from the given options is:

[tex]\[ 1850 \ \text{mL} \][/tex]

Therefore, the correct answer is [tex]\(1850 \ \text{mL}\)[/tex].

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