Answer:
[tex]m_{Al_2O_3}=118.27gAl_2O_3[/tex]
Explanation:
Hello there!
In this case, if we consider the following chemical reaction, whereby Al2O3 is produced from Al and FeO:
[tex]3FeO+2Al\rightarrow 3Fe+Al_2O_3[/tex]
Thus, since there is 3:1 mole ratio of FeO to Al2O3, it turns out feasible for us to use their molar masses, 71.844 g/mol and 101.96 g/mol respectively, to obtain the grams of the latter as follows:
[tex]m_{Al_2O_3}=250.gFeO*\frac{1molFeO}{71.844gFeO}*\frac{1molAl_2O_3}{3molFeO} *\frac{101.96gAl_2O_3}{1molAl_2O_3}\\\\m_{Al_2O_3}=118.27gAl_2O_3[/tex]
Regards!
