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Sagot :
1. The first thing we do is chech if the reaction is balanced:
[tex]2Al\text{ + 3}CuCl_2\rightarrow3Cu+2AlCl_3[/tex]As we can see the equation is balanced because there are on both sides:
2 atoms of Al
3 atoms of Cu
6 atoms of Cl
2. Now we calculate the moles of each reactant that we have:
For this we need to know the molar mass of each reactant. We calculate it using the periodic table.
[tex]\begin{gathered} M_{Al}=26.982\text{ }\frac{g}{mol} \\ M_{CuCl_2}=63.546\text{ }\frac{g}{mol}\text{ + }2\text{ }x\text{ }35.45\frac{g}{mol}=134.446\frac{g}{mol} \end{gathered}[/tex]So we calculate:
[tex]\begin{gathered} n_{Al}=\frac{20g}{26.982\frac{g}{mol}}=0.74\text{ mol} \\ n_{CuCl_2}=\frac{90g}{134.446\frac{g}{mol}}=0.66mol \end{gathered}[/tex]3. Now we calculate the limiting reactant:
If we consider that all the aluminum reacts, for 0.74 moles we would need:
[tex]n_{CuCl_2}=0.669\text{ }mol\text{ }x\frac{3\text{ }molCuCl_2}{2Al\text{ }}=1.11molCuCl_2[/tex]So we see that we need 1.11 moles of CuCl2 in order for the aluminum to react completely. So the limiting reactant is CuCl2.
Now we calculate the amount of copper produced:
So the amount of copper produced is 0.669 moles.
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