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GREYUIT [131]
3 years ago
9

If you had excess chlorine, how many moles of of aluminum chloride could be produced from 11.0 g of aluminum

Chemistry
1 answer:
dlinn [17]3 years ago
5 0

Answer: 0.407 moles of aluminum chloride could be produced from 11.0 g of aluminum.

Explanation:

To calculate the moles :

\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}    

\text{Moles of aluminium}=\frac{11.0g}{27g/mol}=0.407moles

2Al+3Cl_2\rightarrow 2AlCl_3  

According to stoichiometry :

Al is the limiting reagent as it limits the formation of product and Cl_2 is the excess reagent.

As 2 moles of Al give = 2 moles of AlCl_3

Thus moles of Al give =\frac{2}{2}\times 0.407=0.407moles  of AlCl_3

Thus 0.407 moles of aluminum chloride could be produced from 11.0 g of aluminum.

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