To find the maximum amount of magnesium oxide (\( MgO \)) that can be produced during the reaction where 3.2 grams of magnesium (\( Mg \)) reacts with 12.0 grams of oxygen (\( O_2 \)), follow these steps:
1. Determine the molar masses of the reactants:
- Magnesium (\( Mg \)): \( 24.305 \) g/mol
- Oxygen (\( O_2 \)): \( 32.00 \) g/mol
2. Calculate the number of moles of each reactant:
- Moles of \( Mg \):
[tex]\[
\text{moles of Mg} = \frac{\text{mass of Mg}}{\text{molar mass of Mg}} = \frac{3.2 \, \text{g}}{24.305 \, \text{g/mol}} \approx 0.1317 \, \text{mol}
\][/tex]
- Moles of \( O_2 \):
[tex]\[
\text{moles of O_2} = \frac{\text{mass of O_2}}{\text{molar mass of O_2}} = \frac{12.0 \, \text{g}}{32.00 \, \text{g/mol}} = 0.375 \, \text{mol}
\][/tex]
3. Determine the limiting reactant:
- According to the balanced equation: \( 2 \, Mg + O_2 \rightarrow 2 \, MgO \), the molar ratio of \( Mg \) to \( O_2 \) is 2:1.
- Moles of \( O_2 \) needed for the given moles of \( Mg \):
[tex]\[
\text{moles of O_2 needed} = \frac{\text{moles of Mg}}{2} = \frac{0.1317}{2} = 0.0658 \, \text{mol}
\][/tex]
- We have 0.375 moles of \( O_2 \), which is more than the 0.0658 moles needed. Therefore, \( Mg \) is the limiting reactant.
4. Calculate the maximum amount of \( MgO \) produced:
- Since \( Mg \) is the limiting reactant, the moles of \( MgO \) produced will be equal to the moles of \( Mg \):
[tex]\[
\text{moles of MgO produced} = \text{moles of Mg} = 0.1317 \, \text{mol}
\][/tex]
5. Convert the moles of \( MgO \) to grams:
- Molar mass of \( MgO \): \( 40.305 \, \text{g/mol} \)
- Mass of \( MgO \) produced:
[tex]\[
\text{mass of MgO} = \text{moles of MgO} \times \text{molar mass of MgO} = 0.1317 \, \text{mol} \times 40.305 \, \text{g/mol} \approx 5.31 \, \text{g}
\][/tex]
Thus, the maximum amount of magnesium oxide that can be produced during the reaction is approximately \( 5.3 \, \text{grams} \).
The closest answer choice is:
- [tex]\( 5.3 \, \text{grams} \)[/tex]
