Answer:
[tex]n_{Al}=6.00molAl\\\\n_{S}=9.00molS\\\\n_{O}=36.0molO[/tex]
Explanation:
Hello!
In this case, when analyzing the moles of an element inside a compound, we need to keep in mind that a mole ratio should be set up; thus, for aluminum sulfate, we have the following ones:
[tex]\frac{2molAl}{1molAl_2(SO_4)_3} \\\\\frac{3molS}{1molAl_2(SO_4)_3} \\\\\frac{12molO}{1mol1molAl_2(SO_4)_3}[/tex]
Thus, starting by 3.00 moles of aluminum sulfate, the moles of each element turn out:
[tex]n_{Al}=3.00molAl_2(SO_4)_3*\frac{2molAl}{1molAl_2(SO_4)_3} =6.00molAl\\\\n_{S}=3.00molAl_2(SO_4)_3*\frac{3molS}{1molAl_2(SO_4)_3} =9.00molS\\\\n_{O}=3.00molAl_2(SO_4)_3*\frac{12molO}{1molAl_2(SO_4)_3}=36.0molO[/tex]
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