Question

Cate grame de aluminiu de puritate 90% pot arde in 3 moli de oxigen ?

Answer 1

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

$\frac{90}{108} = \frac{100}{x} -\ \textgreater \ x= \frac{108*100}{90} =120$

x=120 g Al 90%

Sper ca te-am ajutat!
Bafta!