Respuesta :
Answer
(a) 172.848 grams
(b) 168.584 grams
Explanation
The balanced equation for the reaction of aluminum with chlorine gas is:
[tex]2Al+3Cl_2\rightarrow2AlCl_3[/tex](a) Mass of Al = 35 g
From the Periodic Table;
Molar mass of Al = 27 g/mol
Molar mass of AlCl₃ = 133.34 g/mol
From the balanced equation above; 2 moles Al produced 2 moles AlCl₃
So in grams, 2 mol x 27 g/mol = 54 g Al produced 2 mol x 133.34 g/mol = 266.68 g AlCl₃
Therefore, 35 g Al will form:
[tex]\frac{266.68\text{ g AlCl}_{3}\text{ }\times35\text{ g Al}}{54\text{ g Al}}=172.848\text{ g AlCl}_{3}[/tex]172.848 grams of aluminum chloride will be formed.
(b) Mass of Al = 42.8 g
Also from the Periodic Table; the molar mass of Clâ‚‚ = 70.90 g/mol
From the equation above; 2 mol of Al requires 3 mol Clâ‚‚.
In grams, 2 mol x 27 g/mol = 54 g Al requires 3 mol x 70.90 g/mol = 212.7 g Clâ‚‚
Hence, 42.8 g Al will react with:
[tex]\frac{42.8\text{ g Al }\times212.7\text{ g Cl}_{2}}{54\text{ g Al}}=168.584\text{ g Cl}_{2}[/tex]168.584 grams of chlorine will react completely with 42.8 g of aluminum
