Answer:
Cl₂ the limiting reactant.
6.67g of AlCl₃ can be produced.
83.5% is percent yield
Explanation:
Based on the reaction:
2 Al + 3Cl₂ → 2AlCl₃
2 moles of aluminium react with 3 moles of chlorine.
To solve this question we must find the moles of each reactant in order to find limiting reactant. With limiting reactant we can find theoretical yield. Percent yield is:
Actual yield (5.57g) / Theoretical yield * 100
That means if we find theoretical yield we can find percent yield:
Moles Aluminium: 26.98g/mol
3.11g * (1mol / 26.98g) = 0.115 moles Al
Moles Chlorine: 70.90g/mol
5.32g * (1mol / 70.90g) = 0.075 moles Cl₂
For a complete reaction of 0.075 moles of Cl₂ are required:
0.075 moles Cl₂ * (2mol Al / 3mol Cl₂) = 0.050 moles of Al
As there are 0.115 moles of Al, Aluminium is the excess reactant and Cl₂ the limiting reactant.
Moles AlCl₃ and mass: 133.34g/mol
0.075 moles Cl₂ * (2mol AlCl₃ / 3mol Cl₂) = 0.050 moles of AlCl₃
0.050 moles of AlCl₃ * (133.34g / mol) =
Percent yield:
5.57g / 6.67g * 100 =
