Odpowiedź:
4Li+O2->2Li2O
n(Li)=0,68mol
n(O2)=(1/4)*n(Li)=(1/4)*0,68=0,17mol
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Odpowiedź:
4Li+O2->2Li2O
n(Li)=0,68mol
n(O2)=(1/4)*n(Li)=(1/4)*0,68=0,17mol