Answer:
1.5 mole
Explanation:
Step 1:
The balanced equation for the reaction. This is illustrated below:
2Al(s) + 3Cl2(g) --> 2AlCl3(s)
Step 2:
Determination of the masses of Al and Cl2 that reacted 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
From the balanced equation,
54g of Al reacted.
213g of Cl2 reacted
Step 3:
Determination of the limiting reactant.
This is illustrated below:
From the balanced equation above,
54g of Al reacted with 213g of Cl2.
Therefore, 40.5g of Al will react with = (40.5 x 213)/54 = 159.75g of Cl2.
From the calculations made above, there are leftover of Cl2 as 159.75g reacted out of 212.7g. Therefore, Cl2 is the excess reactant and Al is the limiting reactant.
Step 4:
Determination of the number of mole in 40.5g of Al. This is illustrated below:
Molar Mass of Al = 27g/mol
Mass of Al = 40.5g
Number of mole of Al =?
Number of mole = Mass/Molar Mass
Number of mole of Al = 40.5/27
Number of mole of Al = 1.5 mole
Step 5:
Determination of the number of mole of AlCl3 produced When 40.5 g of Al and 212.7 g of Cl2 combine together. This is illustrated below:
2Al(s) + 3Cl2(g) --> 2AlCl3(s)
From the balanced equation above,
2 moles of Al produced 2 moles of AlCl3.
Therefore, 1.5 mole of Al will also produce 1.5 mole of AlCl3.
From the calculations made above, 1.5 mole of AlCl3 is produced When 40.5 g of Al and 212.7 g of Cl2 combine together.
