Answer :
Sure! Let's solve this step-by-step to find the percent yield of silver in the reaction.
1. Identify the Reaction:
The balanced chemical equation is:
[tex]\[ \text{Cu} + 2 \text{AgNO}_3 \rightarrow \text{Cu(NO}_3\text{)}_2 + 2 \text{Ag} \][/tex]
2. Given Data:
- Mass of Cu = 12.7 g
- Actual mass of Ag produced = 38.1 g
3. Molar Masses:
- Molar mass of Cu (Copper) = 63.55 g/mol
- Molar mass of Ag (Silver) = 107.87 g/mol
4. Calculate Moles of Copper:
[tex]\[ \text{Moles of Cu} = \frac{\text{Mass of Cu}}{\text{Molar mass of Cu}} \][/tex]
[tex]\[ \text{Moles of Cu} = \frac{12.7 \text{ g}}{63.55 \text{ g/mol}} \approx 0.1998 \text{ moles} \][/tex]
5. Determine Moles of Silver Theoretically Produced:
According to the balanced equation, 1 mole of Cu produces 2 moles of Ag.
[tex]\[ \text{Moles of Ag (theoretical)} = 2 \times \text{Moles of Cu} \][/tex]
[tex]\[ \text{Moles of Ag (theoretical)} = 2 \times 0.1998 \approx 0.3997 \text{ moles} \][/tex]
6. Calculate Theoretical Mass of Silver:
[tex]\[ \text{Theoretical Mass of Ag} = \text{Moles of Ag (theoretical)} \times \text{Molar mass of Ag} \][/tex]
[tex]\[ \text{Theoretical Mass of Ag} = 0.3997 \text{ moles} \times 107.87 \text{ g/mol} \approx 43.11 \text{ g} \][/tex]
7. Calculate Percent Yield:
[tex]\[ \text{Percent Yield} = \left( \frac{\text{Actual Mass of Ag}}{\text{Theoretical Mass of Ag}} \right) \times 100 \][/tex]
[tex]\[ \text{Percent Yield} = \left( \frac{38.1 \text{ g}}{43.11 \text{ g}} \right) \times 100 \approx 88.4\% \][/tex]
So, the percent yield of silver in this reaction is [tex]\( 88.4\% \)[/tex].
Thus, the correct answer is:
[tex]\[ 88.4\% \][/tex]
1. Identify the Reaction:
The balanced chemical equation is:
[tex]\[ \text{Cu} + 2 \text{AgNO}_3 \rightarrow \text{Cu(NO}_3\text{)}_2 + 2 \text{Ag} \][/tex]
2. Given Data:
- Mass of Cu = 12.7 g
- Actual mass of Ag produced = 38.1 g
3. Molar Masses:
- Molar mass of Cu (Copper) = 63.55 g/mol
- Molar mass of Ag (Silver) = 107.87 g/mol
4. Calculate Moles of Copper:
[tex]\[ \text{Moles of Cu} = \frac{\text{Mass of Cu}}{\text{Molar mass of Cu}} \][/tex]
[tex]\[ \text{Moles of Cu} = \frac{12.7 \text{ g}}{63.55 \text{ g/mol}} \approx 0.1998 \text{ moles} \][/tex]
5. Determine Moles of Silver Theoretically Produced:
According to the balanced equation, 1 mole of Cu produces 2 moles of Ag.
[tex]\[ \text{Moles of Ag (theoretical)} = 2 \times \text{Moles of Cu} \][/tex]
[tex]\[ \text{Moles of Ag (theoretical)} = 2 \times 0.1998 \approx 0.3997 \text{ moles} \][/tex]
6. Calculate Theoretical Mass of Silver:
[tex]\[ \text{Theoretical Mass of Ag} = \text{Moles of Ag (theoretical)} \times \text{Molar mass of Ag} \][/tex]
[tex]\[ \text{Theoretical Mass of Ag} = 0.3997 \text{ moles} \times 107.87 \text{ g/mol} \approx 43.11 \text{ g} \][/tex]
7. Calculate Percent Yield:
[tex]\[ \text{Percent Yield} = \left( \frac{\text{Actual Mass of Ag}}{\text{Theoretical Mass of Ag}} \right) \times 100 \][/tex]
[tex]\[ \text{Percent Yield} = \left( \frac{38.1 \text{ g}}{43.11 \text{ g}} \right) \times 100 \approx 88.4\% \][/tex]
So, the percent yield of silver in this reaction is [tex]\( 88.4\% \)[/tex].
Thus, the correct answer is:
[tex]\[ 88.4\% \][/tex]
