Answer:
a)Cl₂ is the limiting reaction
b) Maximum amount of AlCl₃ = 0.45g
Explanation:
2Al(s) + 3Cl2(g) → 2AlCl3(s)
54g 212.72g 266.7g
0.64g 0.36
a) 54g Al ------------ 212.72g
0.64g ------------- ?
mass of cl₂ required
[tex]= \frac{0.64 \times 212.72 }{54} \\\\= 2.52g[/tex]
we have 0.36g Cl₂, so Cl₂ is the limiting reaction
b)
212.72g Cl₂---------- 266.7g AlCl₃
0.36g Cl₂ ----------- ?
mass of AlCl₃ =
[tex]\frac{0.36 \times 266.7 }{212.72} \\\\= 0.45g[/tex]
Maximum amount of AlCl₃ = 0.45g
Answer:
a) Limiting reactant: Cl₂
b) 0.45 g
Explanation:
(a) The balanced equation is:
2Al(s) + 3Cl₂(g) ⇆ 2AlCl₃(s) (1)
0.64g 0.36g
To find the limiting reactant we need to calculate the number of moles of Al and Cl₂:
[tex] \eta_{Al} = \frac{m}{M} [/tex]
where m: is the mass and M: is the molar mass
[tex] \eta_{Al} = \frac{0.64}{27 g/mol} = 0.024 moles [/tex]
Similarly, for Cl₂ we have:
[tex] \eta_{Cl_{2}} = \frac{m}{M} = \frac{0.36 g}{(35.45 g/mol)*2} = 5.07 \cdot 10^{-3} moles [/tex]
From equation (1) we have that 2 moles of Al reacts with 3 moles of Cl₂, hence, we can find the number of moles of Al that reacts with Cl₂ as follows:
2 moles Al → 3 moles Cl₂
0.024 moles Al → x
[tex] x = \frac{0.024 moles Al*3 moles Cl_{2}}{2 moles Al} = 0.036 moles Cl_{2} [/tex]
Hence, we need 0.036 moles of Cl₂ to react with Al, and we have 5.07x10⁻³ moles, so, the limiting reactive is Cl₂ since Al is in excess.
(b) The maximum amount of AlCl₃ in grams that can be produced can be calculated using the limiting reactant, that is to say, Cl₂:
From equation (1) we have that 3 moles of Cl₂ produces 2 moles of AlCl₃, therefore the moles of AlCl₃ are:
[tex]\eta_{AlCl_{3}} = \frac{2 moles AlCl_{3}*5.07 \cdot 10^{-3} moles Cl_{2}}{3 moles Cl_{2}} = 3.38 \cdot 10^{-3} moles[/tex]
Finally, the mass of AlCl₃ is:
[tex]m = mol*M = 3.38 \cdot 10^{-3} moles*133.5g/mol = 0.45 g[/tex]
Therefore, the maximum amount of AlCl₃ that can be produced is 0.45 g.
I hope it helps you!
