Welcome to Westonci.ca, the place where your questions find answers from a community of knowledgeable experts. Connect with professionals on our platform to receive accurate answers to your questions quickly and efficiently. Discover in-depth answers to your questions from a wide network of professionals on our user-friendly Q&A platform.
Sagot :
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
We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. Keep exploring Westonci.ca for more insightful answers to your questions. We're here to help.
