4Al+3O2->2Al2O3
M Al=27 g/mol
4 moli Al.....................3 moli O2
4*27 g Al.....................3 moli O2
90..............100
4*27............x
[tex] \frac{90}{108} = \frac{100}{x} -\ \textgreater \ x= \frac{108*100}{90} =120[/tex]
x=120 g Al 90%
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