Answer :
To determine the volume of oxygen gas required to react with a given mass of magnesium at standard temperature and pressure (STP), we need to follow several steps and use some useful relationships between mass, moles, and volume:
### Step 1: Determine the Moles of Magnesium
First, we need to calculate how many moles of magnesium (\( \text{Mg} \)) we have. We can do this using the formula:
[tex]\[ \text{moles of Mg} = \frac{\text{mass of Mg}}{\text{molar mass of Mg}} \][/tex]
We are given:
- The mass of \( \text{Mg} \) is \( 4.03 \, \text{g} \).
- The molar mass of \( \text{Mg} \) is \( 24.305 \, \text{g/mol} \).
Therefore,
[tex]\[ \text{moles of Mg} = \frac{4.03 \, \text{g}}{24.305 \, \text{g/mol}} = 0.16581 \, \text{mol} \][/tex]
### Step 2: Convert Moles of Magnesium to Moles of Oxygen Gas
According to the balanced chemical equation:
[tex]\[ 2 \, \text{Mg} + \text{O}_2 \rightarrow 2 \, \text{MgO} \][/tex]
From the equation, we see that 2 moles of \( \text{Mg} \) react with 1 mole of \( \text{O}_2 \). Hence, the moles of \( \text{O}_2 \) needed are half the number of moles of \( \text{Mg} \).
Therefore,
[tex]\[ \text{moles of O}_2 = \frac{\text{moles of Mg}}{2} = \frac{0.16581 \, \text{mol}}{2} = 0.08290 \, \text{mol} \][/tex]
### Step 3: Calculate the Volume of \( \text{O}_2 \) Gas at STP
At standard temperature and pressure (STP, where the temperature is \( 0^\circ \text{C} \) and the pressure is \( 1 \, \text{atm} \)), 1 mole of any ideal gas occupies \( 22.414 \, \text{L} \).
Using the relationship:
[tex]\[ \text{volume of O}_2 \, (\text{L}) = \text{moles of O}_2 \times 22.414 \, \text{L/mol} \][/tex]
We get:
[tex]\[ \text{volume of O}_2 \, (\text{L}) = 0.08290 \, \text{mol} \times 22.414 \, \text{L/mol} = 1.8582 \, \text{L} \][/tex]
### Step 4: Convert the Volume from Liters to Milliliters
Finally, we need to convert the volume from liters to milliliters since 1 liter is 1000 milliliters:
[tex]\[ \text{volume of O}_2 \, (\text{mL}) = 1.8582 \, \text{L} \times 1000 \, \text{mL/L} = 1858.2 \, \text{mL} \][/tex]
### Conclusion
The volume of [tex]\( \text{O}_2 \)[/tex] gas required to react with [tex]\( 4.03 \, \text{g} \)[/tex] of [tex]\( \text{Mg} \)[/tex] at STP is approximately [tex]\( 1858.2 \, \text{mL} \)[/tex]. Hence, the correct answer from the given choices is closest to high precision: [tex]\( 1860 \, \text{mL} \)[/tex].
### Step 1: Determine the Moles of Magnesium
First, we need to calculate how many moles of magnesium (\( \text{Mg} \)) we have. We can do this using the formula:
[tex]\[ \text{moles of Mg} = \frac{\text{mass of Mg}}{\text{molar mass of Mg}} \][/tex]
We are given:
- The mass of \( \text{Mg} \) is \( 4.03 \, \text{g} \).
- The molar mass of \( \text{Mg} \) is \( 24.305 \, \text{g/mol} \).
Therefore,
[tex]\[ \text{moles of Mg} = \frac{4.03 \, \text{g}}{24.305 \, \text{g/mol}} = 0.16581 \, \text{mol} \][/tex]
### Step 2: Convert Moles of Magnesium to Moles of Oxygen Gas
According to the balanced chemical equation:
[tex]\[ 2 \, \text{Mg} + \text{O}_2 \rightarrow 2 \, \text{MgO} \][/tex]
From the equation, we see that 2 moles of \( \text{Mg} \) react with 1 mole of \( \text{O}_2 \). Hence, the moles of \( \text{O}_2 \) needed are half the number of moles of \( \text{Mg} \).
Therefore,
[tex]\[ \text{moles of O}_2 = \frac{\text{moles of Mg}}{2} = \frac{0.16581 \, \text{mol}}{2} = 0.08290 \, \text{mol} \][/tex]
### Step 3: Calculate the Volume of \( \text{O}_2 \) Gas at STP
At standard temperature and pressure (STP, where the temperature is \( 0^\circ \text{C} \) and the pressure is \( 1 \, \text{atm} \)), 1 mole of any ideal gas occupies \( 22.414 \, \text{L} \).
Using the relationship:
[tex]\[ \text{volume of O}_2 \, (\text{L}) = \text{moles of O}_2 \times 22.414 \, \text{L/mol} \][/tex]
We get:
[tex]\[ \text{volume of O}_2 \, (\text{L}) = 0.08290 \, \text{mol} \times 22.414 \, \text{L/mol} = 1.8582 \, \text{L} \][/tex]
### Step 4: Convert the Volume from Liters to Milliliters
Finally, we need to convert the volume from liters to milliliters since 1 liter is 1000 milliliters:
[tex]\[ \text{volume of O}_2 \, (\text{mL}) = 1.8582 \, \text{L} \times 1000 \, \text{mL/L} = 1858.2 \, \text{mL} \][/tex]
### Conclusion
The volume of [tex]\( \text{O}_2 \)[/tex] gas required to react with [tex]\( 4.03 \, \text{g} \)[/tex] of [tex]\( \text{Mg} \)[/tex] at STP is approximately [tex]\( 1858.2 \, \text{mL} \)[/tex]. Hence, the correct answer from the given choices is closest to high precision: [tex]\( 1860 \, \text{mL} \)[/tex].
