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Answer :
To determine the percent yield of the reaction where 5.00 grams of aluminum react with 5.00 grams of chlorine gas to form aluminum chloride, and where the actual yield of aluminum chloride is 0.50 grams, we need to go through a series of steps encompassing stoichiometry and yield calculations.
### Step-by-Step Solution:
1. Determine the molar masses:
- Aluminum (Al): 26.98 g/mol
- Chlorine (Cl₂): 35.453 g/mol (for one Cl atom, since chlorine gas is Cl₂, the molar mass for Cl₂ is 2 35.453 = 70.906 g/mol)
- Aluminum chloride (AlCl₃): 26.98 (Al) + 3 35.453 (Cl) = 133.34 g/mol
2. Calculate the moles of aluminum and chlorine:
- Moles of aluminum: [tex]\( \text{moles of Al} = \frac{\text{mass of Al}}{\text{molar mass of Al}} = \frac{5.00 \text{ g}}{26.98 \text{ g/mol}} = 0.1853 \text{ moles} \)[/tex]
- Moles of chlorine: [tex]\( \text{moles of Cl₂} = \frac{\text{mass of Cl₂}}{\text{molar mass of Cl₂}} = \frac{5.00 \text{ g}}{70.906 \text{ g/mol}} = 0.0705 \text{ moles} \)[/tex]
3. Identify the limiting reagent:
The balanced chemical equation for the reaction is:
[tex]\[
2Al + 3Cl₂ → 2AlCl₃
\][/tex]
According to the stoichiometric coefficients:
- 2 moles of Al react with 3 moles of Cl₂.
Thus, we can calculate the moles of aluminum needed to react with the given moles of chlorine:
- Moles of aluminum needed: [tex]\( \text{moles of Cl₂} \times \frac{2 \text{ moles of Al}}{3 \text{ moles of Cl₂}} = 0.0705 \text{ moles of Cl₂} \times \frac{2}{3} = 0.047 \text{ moles of Al} \)[/tex]
Since we have more moles of aluminum (0.1853 moles) than required (0.047 moles), chlorine is the limiting reagent.
4. Calculate the theoretical yield of aluminum chloride:
Based on the limiting reagent (chlorine):
- According to the stoichiometric ratio, 3 moles of Cl₂ produce 2 moles of AlCl₃. Therefore:
[tex]\[ \text{moles of AlCl₃} = \text{moles of Cl₂} \times \frac{2 \text{ moles of AlCl₃}}{3 \text{ moles of Cl₂}} = 0.0705 \text{ moles of Cl₂} \times \frac{2}{3} = 0.047 \text{ moles of AlCl₃} \][/tex]
- The mass of theoretical aluminum chloride:
[tex]\[ \text{mass of AlCl₃} = \text{moles of AlCl₃} \times \text{molar mass of AlCl₃} = 0.047 \text{ moles} \times 133.34 \text{ g/mol} = 6.266 \text{ grams} \][/tex]
5. Calculate the percent yield:
- Percent yield is calculated as:
[tex]\[ \text{Percent yield} = \left( \frac{\text{actual yield}}{\text{theoretical yield}} \right) \times 100 = \left( \frac{0.50 \text{ g}}{6.266 \text{ g}} \right) \times 100 = 8.0\% \][/tex]
So, the percent yield of the reaction given the conditions stated is [tex]\( \boxed{8.0\%} \)[/tex].
### Step-by-Step Solution:
1. Determine the molar masses:
- Aluminum (Al): 26.98 g/mol
- Chlorine (Cl₂): 35.453 g/mol (for one Cl atom, since chlorine gas is Cl₂, the molar mass for Cl₂ is 2 35.453 = 70.906 g/mol)
- Aluminum chloride (AlCl₃): 26.98 (Al) + 3 35.453 (Cl) = 133.34 g/mol
2. Calculate the moles of aluminum and chlorine:
- Moles of aluminum: [tex]\( \text{moles of Al} = \frac{\text{mass of Al}}{\text{molar mass of Al}} = \frac{5.00 \text{ g}}{26.98 \text{ g/mol}} = 0.1853 \text{ moles} \)[/tex]
- Moles of chlorine: [tex]\( \text{moles of Cl₂} = \frac{\text{mass of Cl₂}}{\text{molar mass of Cl₂}} = \frac{5.00 \text{ g}}{70.906 \text{ g/mol}} = 0.0705 \text{ moles} \)[/tex]
3. Identify the limiting reagent:
The balanced chemical equation for the reaction is:
[tex]\[
2Al + 3Cl₂ → 2AlCl₃
\][/tex]
According to the stoichiometric coefficients:
- 2 moles of Al react with 3 moles of Cl₂.
Thus, we can calculate the moles of aluminum needed to react with the given moles of chlorine:
- Moles of aluminum needed: [tex]\( \text{moles of Cl₂} \times \frac{2 \text{ moles of Al}}{3 \text{ moles of Cl₂}} = 0.0705 \text{ moles of Cl₂} \times \frac{2}{3} = 0.047 \text{ moles of Al} \)[/tex]
Since we have more moles of aluminum (0.1853 moles) than required (0.047 moles), chlorine is the limiting reagent.
4. Calculate the theoretical yield of aluminum chloride:
Based on the limiting reagent (chlorine):
- According to the stoichiometric ratio, 3 moles of Cl₂ produce 2 moles of AlCl₃. Therefore:
[tex]\[ \text{moles of AlCl₃} = \text{moles of Cl₂} \times \frac{2 \text{ moles of AlCl₃}}{3 \text{ moles of Cl₂}} = 0.0705 \text{ moles of Cl₂} \times \frac{2}{3} = 0.047 \text{ moles of AlCl₃} \][/tex]
- The mass of theoretical aluminum chloride:
[tex]\[ \text{mass of AlCl₃} = \text{moles of AlCl₃} \times \text{molar mass of AlCl₃} = 0.047 \text{ moles} \times 133.34 \text{ g/mol} = 6.266 \text{ grams} \][/tex]
5. Calculate the percent yield:
- Percent yield is calculated as:
[tex]\[ \text{Percent yield} = \left( \frac{\text{actual yield}}{\text{theoretical yield}} \right) \times 100 = \left( \frac{0.50 \text{ g}}{6.266 \text{ g}} \right) \times 100 = 8.0\% \][/tex]
So, the percent yield of the reaction given the conditions stated is [tex]\( \boxed{8.0\%} \)[/tex].
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