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Sagot :
The first step to solve this question is to convert the given masses to moles using the corresponding molecular weight:
[tex]15.2gAl\cdot\frac{molAl}{27gAl}=0.56molAl[/tex][tex]39.1gCl\cdot\frac{molCl_2}{35.45gCl_2}=1.1molCl_2[/tex]Divide each of the results by the coefficient of each reactant in the equation. It means, divide the amount of Al by 2 and the amount of Cl2 by 3:
[tex]\begin{gathered} \frac{0.56molAl}{2molAl}=0.28 \\ \frac{1.1molCl_2}{3molCl_2}=0.367 \end{gathered}[/tex]It means that the limiting reactant is Aluminium and we have to base our calculations on this reactant. From the equation we know than 2 moles of Al produce 2 moles of AlCl3.
[tex]0.56molAl\cdot\frac{2molAlCl_3}{2molAl}=0.56molAlCl_3[/tex]Now, convert this amount of AlCl3 to mass using its molecular weight:
[tex]0.56molAlCl_3\cdot\frac{133.33gAlCl_3}{molAlCl_3}=74.66gAlCl_3[/tex]It means that the mass of AlCl3 formed is 74.66g.
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