Answer:
A. Chlorine (Cl2) is the limiting reactant.
B. 6.77g of AlCl3
Step-by-step explanation:
The balanced equation for the reaction. This is given below:
2Al + 3Cl2 —> 2AlCl3
Next, we shall determine the masses of Al and Cl2 that reacted and the mass of AlCl3 produced from the balanced equation. This is illustrated below:
Molar Mass of Al = 27g/mol
Mass of Al from the balanced equation = 2 x 27 = 54g
Molar Mass of Cl2 = 2 x 35.5 = 71g/mol
Mass of Cl2 from the balanced equation = 3 x 71 = 213g
Molar Mass of AlCl3 = 27 + (3x35.5) = 133.5g/mol
Mass of AlCl3 from the balanced equation = 2 x 133.5 = 267g
Thus, from the balanced equation above,
54g of Al reacted with 213g of Cl2 to produced 267g of AlCl3.
A. Determination of the limiting reactant. This is illustrated below:
From the balanced equation above,
54g of Al reacted with 213g of Cl2.
Therefore, 3.2g of Al will react will react with = (3.2 x 213)/54 = 12.52g of Cl2.
We can see that it will take a higher mass of Cl2 ( i.e 12.52g) than what was given ( i.e 5.4g) to react with 3.2g of Al.
Therefore, Cl2 is the limiting reactant.
B. Determination of the mass of AlCl3 produced from the amount of the limiting reactant available.
This is illustrated below:
The limiting reactant is Cl2 with a mass of 5.4g
From the balanced equation above, 213g of Cl2 reacted to produced 267g of AlCl3.
Therefore, 5.4g of Cl2 will react to produce = (5.4 x 267)/213 = 6.77g of AlCl3.
Therefore, 6.77g of AlCl3 is produced from the amount of the limiting reactant available.