Answer:
[tex]m_{Al_2(SO_4)_3}=17.5gAl_2(SO_4)_3[/tex]
Explanation:
Hello,
In this case, the undergoing balanced chemical reaction is:
[tex]Al_2O_3(s)+3H_2SO_4(aq)\rightarrow Al_2(SO_4)_3(aq)+3H_2O(l)[/tex]
Thus, as 5.22 grams of aluminium oxide reacts, the required yielded amount of aluminium sulfate results:
[tex]m_{Al_2(SO_4)_3}=5.22gAl_2O_3*\frac{1molAl_2O_3}{102gAl_2O_3}*\frac{1molAl_2(SO_4)_3}{1molAl_2O_3}*\frac{342gAl_2(SO_4)_3}{1molAl_2(SO_4)_3} \\\\m_{Al_2(SO_4)_3}=17.5gAl_2(SO_4)_3[/tex]
Moreover, the percent yield is:
[tex]Y=\frac{12.9g}{17.5g} *100\%=73.7\%[/tex]
Best regards.