In the following reaction 12.0 g of Mg reacts with excess oxygen to produce 12.1 g of MgO. What is the percent yield (in %) in this reaction? 2Mg + O2 → 2MgO Answer: K
In the following reaction 12.0 g of Mg reacts with excess oxygen to produce 12.1 g of MgO. What is the percent yield (in %) in this reaction? 2Mg + O2 → 2MgO Answer: K
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![### Percent Yield Calculation for Magnesium Oxide Reaction
**Question:**
In the following reaction, 12.0 g of Mg reacts with excess oxygen to produce 12.1 g of MgO. What is the percent yield (in %) in this reaction?
\[ 2\text{Mg} + \text{O}_2 \rightarrow 2\text{MgO} \]
**Answer:**
To determine the percent yield, we need to follow these steps:
1. Calculate the moles of Mg used.
2. Use stoichiometry to calculate the theoretical mass of MgO produced.
3. Calculate the percent yield using the actual mass of MgO produced and the theoretical mass calculated.
1. **Calculate the Moles of Mg Used:**
\[ \text{Moles of Mg} = \frac{\text{Mass of Mg}}{\text{Molar mass of Mg}} \]
Given:
- Mass of Mg = 12.0 g
- Molar mass of Mg = 24.305 g/mol
\[ \text{Moles of Mg} = \frac{12.0 \, \text{g}}{24.305 \, \text{g/mol}} = 0.4937 \, \text{mol} \]
2. **Calculate the Theoretical Mass of MgO Produced:**
According to the balanced chemical equation:
\[ 2\text{Mg} + \text{O}_2 \rightarrow 2\text{MgO} \]
- 2 moles of Mg produce 2 moles of MgO.
Since 1 mole of Mg produces 1 mole of MgO, 0.4937 moles of Mg will produce 0.4937 moles of MgO.
The molar mass of MgO is:
- Molar mass of Mg (+) = 24.305 g/mol
- Molar mass of O = 16.00 g/mol
- Molar mass of MgO = 24.305 g/mol + 16.00 g/mol = 40.305 g/mol
Theoretical mass of MgO:
\[ \text{Theoretical mass of MgO} = \text{Moles of MgO} \times \text{Molar mass of MgO} \]
\[ \text{Theoretical mass of MgO} = 0.4937 \, \text{mol} \times](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6ae95cc5-a0cd-43b0-8331-9678c9dacca0%2Ff739afd8-a7be-470f-8e7f-a55e870912e6%2F8ywq35q.jpeg&w=3840&q=75)
Transcribed Image Text:### Percent Yield Calculation for Magnesium Oxide Reaction
**Question:**
In the following reaction, 12.0 g of Mg reacts with excess oxygen to produce 12.1 g of MgO. What is the percent yield (in %) in this reaction?
\[ 2\text{Mg} + \text{O}_2 \rightarrow 2\text{MgO} \]
**Answer:**
To determine the percent yield, we need to follow these steps:
1. Calculate the moles of Mg used.
2. Use stoichiometry to calculate the theoretical mass of MgO produced.
3. Calculate the percent yield using the actual mass of MgO produced and the theoretical mass calculated.
1. **Calculate the Moles of Mg Used:**
\[ \text{Moles of Mg} = \frac{\text{Mass of Mg}}{\text{Molar mass of Mg}} \]
Given:
- Mass of Mg = 12.0 g
- Molar mass of Mg = 24.305 g/mol
\[ \text{Moles of Mg} = \frac{12.0 \, \text{g}}{24.305 \, \text{g/mol}} = 0.4937 \, \text{mol} \]
2. **Calculate the Theoretical Mass of MgO Produced:**
According to the balanced chemical equation:
\[ 2\text{Mg} + \text{O}_2 \rightarrow 2\text{MgO} \]
- 2 moles of Mg produce 2 moles of MgO.
Since 1 mole of Mg produces 1 mole of MgO, 0.4937 moles of Mg will produce 0.4937 moles of MgO.
The molar mass of MgO is:
- Molar mass of Mg (+) = 24.305 g/mol
- Molar mass of O = 16.00 g/mol
- Molar mass of MgO = 24.305 g/mol + 16.00 g/mol = 40.305 g/mol
Theoretical mass of MgO:
\[ \text{Theoretical mass of MgO} = \text{Moles of MgO} \times \text{Molar mass of MgO} \]
\[ \text{Theoretical mass of MgO} = 0.4937 \, \text{mol} \times
Transcribed Image Text:MISSED THIS? Watch KCV: Stoichiometry,
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Hydrobromic acid dissolves solid iron according to
the reaction
Fe(s) + 2HBr(aq) → FeBr2 (aq) + H₂(g)
▼
Part A
What mass of HBr (in g) would you need to dissolve a 3.40-g pure iron bar on a
padlock?
Express your answer in grams to three significant figures.
► View Available Hint(s)
m =
17 ΑΣΦ
?
6.0
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