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To determine the maximum amount of magnesium oxide (\( \text{MgO} \)) that can be produced during a reaction where 3.2 grams of magnesium (\( \text{Mg} \)) react with 12.0 grams of oxygen (\( \text{O}_2 \)), we need to follow a step-by-step stoichiometric calculation based on the balanced chemical equation:
[tex]\[ 2 \text{Mg} + \text{O}_2 \rightarrow 2 \text{MgO} \][/tex]
### Step 1: Calculate the number of moles of each reactant.
Mass of Mg:
Given:
- Mass of Mg = 3.2 grams
- Molar mass of Mg = 24.305 g/mol
Moles of Mg:
[tex]\[ \text{moles of Mg} = \frac{\text{mass of Mg}}{\text{molar mass of Mg}} = \frac{3.2 \text{ g}}{24.305 \text{ g/mol}} \approx 0.13166 \text{ moles} \][/tex]
Mass of \( \text{O}_2 \):
Given:
- Mass of \( \text{O}_2 \) = 12.0 grams
- Molar mass of \( \text{O}_2 \) = 32.00 g/mol
Moles of \( \text{O}_2 \):
[tex]\[ \text{moles of } \text{O}_2 = \frac{\text{mass of } \text{O}_2}{\text{molar mass of } \text{O}_2} = \frac{12.0 \text{ g}}{32.00 \text{ g/mol}} = 0.375 \text{ moles} \][/tex]
### Step 2: Determine the limiting reactant.
According to the balanced equation:
[tex]\[ 2 \text{Mg} + \text{O}_2 \rightarrow 2 \text{MgO} \][/tex]
- 2 moles of Mg react with 1 mole of \( \text{O}_2 \).
Calculate the theoretical yield of \( \text{MgO} \) from each reactant.
#### Using Mg (assuming Mg is the limiting reactant):
- 1 mole of Mg produces 1 mole of \( \text{MgO} \).
[tex]\[ \text{Moles of } \text{MgO} \text{ produced (from Mg)} = \text{moles of Mg} = 0.13166 \text{ moles} \][/tex]
#### Using \( \text{O}_2 \) (assuming \( \text{O}_2 \) is the limiting reactant):
- 1 mole of \( \text{O}_2 \) produces 2 moles of \( \text{MgO} \).
[tex]\[ \text{Moles of } \text{MgO} \text{ produced (from } \text{O}_2 \text{) = 2} \times \text{moles of } \text{O}_2 = 2 \times 0.375 \text{ moles} = 0.75 \text{ moles} \][/tex]
The actual maximum moles of \( \text{MgO} \) produced are determined by the limiting reactant.
Since \( 0.13166 \text{ moles (from Mg)} < 0.75 \text{ moles (from } \text{O}_2 \text{)}\), Mg is the limiting reactant.
### Step 3: Calculate the mass of \( \text{MgO} \) produced.
Given:
- Molar mass of \( \text{MgO} \) = 40.305 g/mol
Mass of \( \text{MgO} \) produced:
[tex]\[ \text{mass of } \text{MgO} = \text{moles of } \text{MgO} \times \text{molar mass of } \text{MgO} = 0.13166 \text{ moles} \times 40.305 \text{ g/mol} \approx 5.31 \text{ grams} \][/tex]
### Step 4: Answer the question.
The maximum amount of magnesium oxide (\( \text{MgO} \)) that can be produced during the reaction is approximately 5.31 grams. Among the given options, this value is closest to 5.3 grams.
Therefore, the correct answer is:
5.3 grams
[tex]\[ 2 \text{Mg} + \text{O}_2 \rightarrow 2 \text{MgO} \][/tex]
### Step 1: Calculate the number of moles of each reactant.
Mass of Mg:
Given:
- Mass of Mg = 3.2 grams
- Molar mass of Mg = 24.305 g/mol
Moles of Mg:
[tex]\[ \text{moles of Mg} = \frac{\text{mass of Mg}}{\text{molar mass of Mg}} = \frac{3.2 \text{ g}}{24.305 \text{ g/mol}} \approx 0.13166 \text{ moles} \][/tex]
Mass of \( \text{O}_2 \):
Given:
- Mass of \( \text{O}_2 \) = 12.0 grams
- Molar mass of \( \text{O}_2 \) = 32.00 g/mol
Moles of \( \text{O}_2 \):
[tex]\[ \text{moles of } \text{O}_2 = \frac{\text{mass of } \text{O}_2}{\text{molar mass of } \text{O}_2} = \frac{12.0 \text{ g}}{32.00 \text{ g/mol}} = 0.375 \text{ moles} \][/tex]
### Step 2: Determine the limiting reactant.
According to the balanced equation:
[tex]\[ 2 \text{Mg} + \text{O}_2 \rightarrow 2 \text{MgO} \][/tex]
- 2 moles of Mg react with 1 mole of \( \text{O}_2 \).
Calculate the theoretical yield of \( \text{MgO} \) from each reactant.
#### Using Mg (assuming Mg is the limiting reactant):
- 1 mole of Mg produces 1 mole of \( \text{MgO} \).
[tex]\[ \text{Moles of } \text{MgO} \text{ produced (from Mg)} = \text{moles of Mg} = 0.13166 \text{ moles} \][/tex]
#### Using \( \text{O}_2 \) (assuming \( \text{O}_2 \) is the limiting reactant):
- 1 mole of \( \text{O}_2 \) produces 2 moles of \( \text{MgO} \).
[tex]\[ \text{Moles of } \text{MgO} \text{ produced (from } \text{O}_2 \text{) = 2} \times \text{moles of } \text{O}_2 = 2 \times 0.375 \text{ moles} = 0.75 \text{ moles} \][/tex]
The actual maximum moles of \( \text{MgO} \) produced are determined by the limiting reactant.
Since \( 0.13166 \text{ moles (from Mg)} < 0.75 \text{ moles (from } \text{O}_2 \text{)}\), Mg is the limiting reactant.
### Step 3: Calculate the mass of \( \text{MgO} \) produced.
Given:
- Molar mass of \( \text{MgO} \) = 40.305 g/mol
Mass of \( \text{MgO} \) produced:
[tex]\[ \text{mass of } \text{MgO} = \text{moles of } \text{MgO} \times \text{molar mass of } \text{MgO} = 0.13166 \text{ moles} \times 40.305 \text{ g/mol} \approx 5.31 \text{ grams} \][/tex]
### Step 4: Answer the question.
The maximum amount of magnesium oxide (\( \text{MgO} \)) that can be produced during the reaction is approximately 5.31 grams. Among the given options, this value is closest to 5.3 grams.
Therefore, the correct answer is:
5.3 grams
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