Be sure to answer all parts. The balanced equation for the reaction of aluminum metal and chlorine gas is 2Al(s) + 3Cl2(g) → 2AlCl3(s) Assume that 0.80 g Al is mixed with 0.23 g Cl2. (a) What is the limiting reactant? Cl2 Al (b) What is the maximum amount of AlCl3, in grams, that can be produced? g AlCl3

Respuesta :

Answer: a) [tex]Cl_2[/tex] is the limiting reagent

b)  0.27 g of [tex]AlCl_3[/tex] will be produced.

Explanation:

To calculate the moles :

[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]    

[tex]\text{Moles of} Al=\frac{0.80g}{27g/mol}=0.030moles[/tex]

[tex]\text{Moles of} Cl_2=\frac{0.23g}{71g/mol}=0.003moles[/tex]

[tex]2Al(s)+3Cl_2(g)\rightarrow 2AlCl_3(s)[/tex]

According to stoichiometry :

3 moles of [tex]Cl_2[/tex] require = 2 moles of [tex]Al[/tex]

Thus 0.003 moles of [tex]Cl_2[/tex] will require=[tex]\frac{2}{3}\times 0.003=0.002moles[/tex]  of [tex]Al[/tex]

Thus [tex]Cl_2[/tex] is the limiting reagent as it limits the formation of product and [tex]Al[/tex] is the excess reagent.

b) As 3 moles of [tex]Cl_2[/tex] give = 2 moles of [tex]AlCl_3[/tex]

Thus 0.003 moles of [tex]Cl_2[/tex] give =[tex]\frac{2}{3}\times 0.003=0.002moles[/tex]  of [tex]AlCl_3[/tex]

Mass of [tex]AlCl_3=moles\times {\text {Molar mass}}=0.002moles\times 133g/mol=0.27g[/tex]

Thus 0.27 g of [tex]AlCl_3[/tex] will be produced.

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