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Sagot :
Answer:
74.9g of AlCl3 are formed.
Explanation:
1st) It is necessary to write and balance the chemical reaction:
[tex]2Al+3Cl_2\rightarrow2AlCl_3[/tex]From the balanced reaction, we know that 2 moles of Al react with 3 moles of Cl2 to produce 2 moles of AlCl3.
2nd) With the molar mass of Al and Cl2, we can convert the moles to grams:
- Al molar mass: 27g/mol
- Al conversion:
[tex]2moles*\frac{27g}{1mole}=54g[/tex]- Cl2 molar mass: 71g/mol
- Cl2 conversion:
[tex]3moles\frac{71g}{1mole}=213g[/tex]Now we know that from the balanced reaction, 54g of Al react with 213g of Cl2.
3rd) Now we have to find out which compound is the limiting reactant and which compound is the excess reactant:
[tex]\begin{gathered} 54gAl-213gCl_2 \\ 15.2gAl-x=\frac{15.2gAl*213gCl_2}{54gAl} \\ x=59.96gCl_2 \end{gathered}[/tex]Now we know that the 15.2g of Al will need 59.96g of Cl2, but we only have 39.1g of Cl2, so Cl2 is the limiting reactant and Al is the excess reactant.
4th) With the molar mass of AlCl3, we can convert the moles to grams of AlCl3:
- AlCl3 molar mass: 133g/mol
- AlCl3 conversion:
[tex]2moles*\frac{133g}{1mole}=266g[/tex]266g of AlCl3 are formed from the balanced reaction.
5th) Finally, with the limiting reactant and the grams of AlCl3 that must be produced from the balanced reaction, we can calculate the grams of AlCl3 that will be formed from 15.2g of aluminum:
[tex]\begin{gathered} 54gAl-266gAlCl_3 \\ 15.2gAl-x=\frac{15.2gAl*266gAlCl_3}{54gAl} \\ x=74.9gAlCl_3 \end{gathered}[/tex]So, 74.9g of AlCl3 are formed.
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