Aluminum and cloride undergoes a synthesis reaction if 8molnof Al reacts with 10mol of cl, what is the maximum amount of AlCl3 can produce
Aluminum and cloride undergoes a synthesis reaction if 8molnof Al reacts with 10mol of cl, what is the maximum amount of AlCl3 can produce
Answer: The mass of [tex]AlCl_3[/tex] produced is 889.38 g
Explanation:
We are given:
Moles of Al = 8 mol
Moles of [tex]Cl_2[/tex] = 10 mol
For the given chemical reaction:
[tex]2Al+3Cl_2\rightarrow 2AlCl_3[/tex]
By stoichiometry of the reaction:
If 3 moles of chlorine gas reacts with 2 moles of Al
So, 10 moles of chlorine gas will react with = [tex]\frac{2}{3}\times 10=6.67mol[/tex] of Al
As the given amount of Al is more than the required amount. Thus, it is present in excess and is considered as an excess reagent.
Thus, chlorine gas is considered a limiting reagent because it limits the formation of the product.
By the stoichiometry of the reaction:
If 3 moles of [tex]Cl_2[/tex] produces 2 mole of [tex]AlCl_3[/tex]
So, 10 moles of [tex]Cl_2[/tex] will produce = [tex]\frac{2}{3}\times 10=6.67mol[/tex] of [tex]AlCl_3[/tex]
The number of moles is defined as the ratio of the mass of a substance to its molar mass. The equation used is:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
We know, molar mass of [tex]AlCl_3[/tex] = 133.34 g/mol
Putting values in above equation, we get:
[tex]\text{Mass of }AlCl_3=(6.67mol\times 133.34g/mol)=889.38g[/tex]
Hence, the mass of [tex]AlCl_3[/tex] produced is 889.38 g