Answer:
[tex]Y=34.3\%[/tex]
Explanation:
Hello there!
In this case, according to the balanced chemical reaction:
[tex]2Al + Cr_2O_3 \rightarrow Al_2O_3 + 2Cr[/tex]
We notice there is a 1:1 mole ratio between Al2O3 and Cr2O3; thus, the following stoichiometric setup is used to compute the theoretical yield first:
[tex]m_{Al_2O_3 }=23.3gCr_2O_3*\frac{1molCr_2O_3}{151.99gCr_2O_3} *\frac{1molAl_2O_3}{1molCr_2O_3} *\frac{101.96gAl_2O_3}{1molAl_2O_3} \\\\m_{Al_2O_3}=15.6gAl_2O_3[/tex]
Thus, the percent yield turns out:
[tex]Y=\frac{5.35g}{15.6g} *100\%\\\\Y=34.3\%[/tex]
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