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The balanced equation for the reaction of aluminum metal and chlorine gas is 2Al(s) + 3Cl2(g) → 2AlCl3(s) Assume that 0.64 g Al is mixed with 0.36 g Cl2. (a) What is the limiting reactant? Al Cl2 (b) What is the maximum amount of AlCl3, in grams, that can be produced?

Respuesta :

Olajidey

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

whitneytr12

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!

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